Amar G. Bose: 6.312 Lecture 13

DR. AMAR G. BOSE: Today we have only one topic-- radiation impedance.

By the way, I just remembered that, driving home Thursday- somehow once in a while you review what you've done, and I recall that we got into a little side discussion of viscosity. And it all of a sudden flashed before me that I wrote something on the board-- I hope I said what's correct, but what I wrote on board was definitely not correct.

And we talked about any situation which had velocity shear, so to speak, that this is the velocity as you go up on the y-axis, and the wave is going this way. The amount of change of the velocity as you go this way is what gives rise to a pressure here to move through a viscous medium, if the medium has viscosity. And the expression is something like this. This is not particularly important for the subject, but I recall that I think I wrote a different thing down. du dy. And I think I wrote dy dx or something.

In other words, the amount of so-called velocity shear is what gives rise to the pressure in this thing is called alternately coefficient of viscosity or just plain viscosity. The bigger the viscosity, obviously, the bigger the pressure for a given average velocity.

So radiation impedance. You what it is to calculate the impedance into two terminals of an electrical network, or a mechanical network. You know what it is to calculate the impedance in looking into a tube, open end or closed, or whatever is hanging on the other end. Radiation impedance is no mystery. It is simply looking out of something at the rest of the world. In other words, the tube that you're able to compute the input impedance, z-- if you looked at the impedance at this point here, this would be a radiation impedance, looking out at the rest of the world. That's all it is.

If it's a specific acoustic impedance, the ratio of pressure, complex amplitudes, of course, for impedance, pressure to velocity at this point is the radiation impedance. It's called radiation only because this is launching the wave into the rest of space. It's called another impedance, input impedance for this tube over here. That's all.

Now, what the radiation impedance looks like in practical devices has a significant role to play on the equipment that you use to drive the devices. And we will see soon that the radiation impedance looking out of, for example, something like this, or something like a piston or a loudspeaker in an infinite baffle-- this will tell us a lot about the problems of how do you get some sound out here? It's harder to get it out of some frequencies than it is at other frequencies. And that's because, looking out here, this fellow sees a different impedance at different frequencies.

Now, you can go ahead and solve the wave equations here to get this radiation impedance. One way, for example, would be to, if this were an infinite baffle to do what we had learned before. Think of this as an array of point sources here, and come out here, and get an expression for the pressure as a function of the space variables. Then assume that this is driven by some u sub s. That gives you enough to calculate the pressure out of here by superposition of a lot of little point sources, an integral which we developed before.

Then when you get that pressure, look at the expression as you go down very close to the surface. That gives you the pressure at each point, which won't necessarily be the same at each point on this-- let's say it's a circular piston. From that, you can add up and get the total force on the piston, and now you have enough to see what the impedance is. If you have the total force on the piston and you know its velocity, you have its mechanical impedance from which you can get all the other impedances is just by area factors.

So that's one way to get it. Another way would be to solve the wave equation up here in space, subject to the boundary condition that the normal velocity 0 along this infinite plane except here and its use of s from here to here. You get the ratio of the pressure to the velocity, or pressure to volume velocity, or force to velocity, whichever impedance you want. And that is the radiation impedance.

Now, funny things happen when you do this. This is going ahead, but we're going to see it in a couple of minutes. When you do this, you don't get something like for impedance of a plane wave going along the x-axis, or a wave travelling down a tube. You see a real impedance if the tube is infinitely long. You don't see that when you look out from practical devices. If you did, we would have a great time in terms of antennas, in terms of loudspeakers. It would be much easier to radiate certain frequencies than it now is.

And there are funny things going on near here, meaning within a few wavelengths of whatever frequency you're driving that cause a lot of energy to be exchanged. It goes out here, and it comes back. It goes out and it comes back. And just like we saw on the power lines, how the power company doesn't like that, because it has to generate a lot of extra current to give you the same power, if you wish. Well, you're going to see why a piston doesn't like that, either. And it's much nicer if all the action here were in-phase, if the velocity and the pressure were in-phase. And then it would be great, Because all the motion would contribute to power-- pressure, a real part of p times u, and that's power that's going, and it doesn't come back. So in that case, it's radiated power.

So what happens in this area nearby here is-- let's say, the area itself is called the near field of a radiating source. You have near fields for electromagnetic antennas. In the very same sense you pump into a dipole or whatever you want, you pump current in here, and you develop a voltage across here. Well, voltage isn't in-phase with the current at some frequencies for any kind of an antenna. And so you have energy exchanging between-- if the voltage isn't in-phase here, we know that that means stored energy, right? And so the fields around that antenna store energy, they exchange it back to the source, and that creates a problem which you have to get around.

In many antenna problems where you're transmitting a radio signal, it's not a big problem, because it's very narrow band, and you can treat it as the modulation of, let's say, plus and minus 15 kilohertz on an FM station that's 100 megahertz. For all practical purposes, the 100 megahertz is only changing by plus and minus 15, so you can tune the antenna so that the impedance is real, looking in here at that frequency, and then you're all set. But when you need something broadband, you can't do it. And if you try in the audio region, 24 decades, 20 to 20 kilohertz, to move a piston there, and you can't tune it to be looking at a real impedance out here, no matter what kind of apparatus you put outside.

So turns out that it's really easy to look at all these effects in terms of a very simple radiator, which embodies the basics of the problems that you find in all the complicated ones, but the math is very simple. And we can get the insight from that, and then you can just carry it over to situations that I talked about here. And that is the pulsating sphere.

Basically, we have a sphere of some radius r0, and what we're going to do is look at the world the way that pulsating sphere does. Let's say he's expanding-- don't ask how. Imagine it's a balloon or something, and its radius is uniformly pulsating in and out with some velocity, and that generates a pressure, and you try to push the air away. The ratio of the pressure to the velocity is a specific acoustic impedance, and we just want to find that. This is kind of-- there's no more mathematics in this than you've already done on your homework.

We know this is now radiating into space, so there's no objects out there, so there's no P minus coming back. So the complex amplitude of the pressure P of r and omega, we can express as P plus over r-- and I'm out here somewhere now, for the moment. My r is out here. P plus over r e to the minus j omega t minus-- sorry. Omega r over c. I put a P plus there just to remind us that the whole P is constituted from a P plus. There is no P minus.

Now we want to find the ratio between pressure and velocity. Well, we don't have velocity, but we know that the velocity, of course, satisfies the same kind of wave equation, and it's going to have a complex amplitude u of r and omega. And we just want to-- since I don't need any time functions here. I'm looking for impedances. So we can just bring in that u of r and omega, and relate it to the P by Newton's Law. And Newton's Law in time said that partial of pressure with respect to x-- t is equal to-- oh, sorry. Is equal to minus rho 0 partial of u with respect to t. u is the velocity. These are instantaneous quantities.

Now, this whole thing, if I let little p be cap P, e to the j omega t, little u be cap u, e to the j omega t-- then of course this time domain expression goes over to, in complex amplitudes, partial of P of r and omega with respect to r is equal to minus rho zero.

Well, when I take the derivative with respect to t, that, of course, in the complex brings down j omega rho 0 and u of r and omega. In other words, u of t is u of r and omega, e to the j omega t. And you put that in and you drop out the j omega. We've done this many times. That's why I did it just by putting it down, almost, here.

OK. So if we have this, we now have a relationship between this and the complex amplitude of the velocity and the ratio of this to this is what we're looking for, the impedance. So all we have to do is use this expression in here, and that will bring us to the relation.

Partial of P of r and omega with respect to r. OK. I take this and I get a minus 1 over r times P-- I'm going to take a little shortcut now, because I've done this, I think, in class once before, and I know you've done it in a homework problem, but if there are questions asked-- P of r and omega. In other words, if I differentiate with respect to this r, it's a product, I get 1 minus 1 over r squared, P plus times all this. But this much of it is P of omega. So it's minus 1 over r times P of r and omega plus or minus-- now the next term, I differentiate that with respect to r, I get j omega over c times the same thing. P of r and omega equals minus-- my minus sign should've been here from here-- minus j omega rho 0 u of r and omega.

OK. All minus signs, they'll go out. So P of r and omega-- let's see if we can do this all at once. Over u of r and omega. Now, by the way, when I say P of r and omega and u of r and omega. I had my sphere here, which was radius r0, and I was out here somewhere a radius r. I'm computing the impedance looking out here at the moment, and until I specialize and say that r in here is equal to r0, which I will do. But I'm just looking at the impedance at any point out here, at the moment.

And this is z of r and omega. I won't call it the radiation impedance yet. It's just the impedance seen by the wave, looking out from here. P divided by u is-- all the minus signs are out. That is 1/r plus j omega over c downstairs and j omega rho 0 upstairs. If I multiply that to get it in a standard form, multiply top and bottom by c, I'll get j omega plus c over r, j omega rho 0 c, which I can express, since rho 0 c is z0, z0 times j omega over j omega plus c over r.

Now, it's interesting to look at this expression. By the way, I can-- I'll do it later. Look what happens to this expression when omega is very large compared to c over r. And by the way, what does that mean? Omega very, very large, compared to c/r, means omega/c, very, very large compared to 1/r. And that means that omega over c is 2 pi/lambda, 2 pi/lambda much, much greater than 1/r, or r much, much greater than lambda/2 pi. Again, lambda/2 pi pops up everywhere.

So if you're out here much, much greater than 6 to the wavelength-- namely a few wavelengths, at least-- what happens is, this term dominates this, and look what happens then. For omega much, much greater-- well, either way you want to put it. For omega much, much greater than c/r, z of r and omega goes to z0.

In other words, when we get far enough away, the spherical wave that's traveling out thinks it's a plane wave. And it thinks that because what really happens, if you look at it-- here's, let's say, a spherical wave going out. This little part here, if I'm computing what the impedance is seen by him, it very much depends on what his neighbors are doing. If his neighbors aren't doing anything-- in other words, if I have a tube here, and I'm looking at the impedance here, and nothing is moving out here, then I see a totally different impedance than if I had another tube here with the same wave going down it, and it was pushing forward, too. So it constrains this thing not to have divergence.

If I had a surface like this-- remember looking at the impedance-- what would happen? It notices that the surface next to it is doing that, and he's radiating like that-- that's compression. And so if I had that surface, I would have an opposite sign. Not looking like a mass, but looking like a compression. If I have a convex surface, the next surfaces around it are radiating out in that direction, and there's divergence. And this fellow now sees a mass out there. He sees a compliance if there's convergence, he sees a mass if there's divergence, and if everything is going straight, he sees z0, real.

So what that little element sees depends upon this. Now, if the sphere is very small, there's a lot of divergence. Those arrows are going way out, and he sees a lot of mass. If it were the other way around, he would see a lot of compression, if the wave was going like so. If it's straight, it's straight.

So you can almost look at a surface, and you can see what kind of impedance that surface is going to see. If, with respect to the wavelength, the thing looks sort of flat, it's real. If it's looking the other way, it's a compression. If it's looking from a convex surface, it's a mass.

Now let's take a look at this at the other end. Namely, if r gets very, very small. r gets very, very small. You know what it's going to be, small compared to lambda/2 pi. But I think you can work that out, too. If it gets very small, this turn gets very large. This one gets neglected. And look what happens. The whole impedance becomes imaginary.

So for r-- well, I said omega here, so I'll say omega much, much less-- yeah. The way I looked at this one was varying frequencies for a fixed r, so let me look at this one the same way. You can choose anything you want, any way you wish. You can choose a fixed frequency and a different r, or a fixed r in different frequency. Since I started this way, I'll do it again this way. For omega much much, much less than c/r, which means that r was much, much less than lanbda/2, if you want to look at it that way. z, this impedance, z of r and omega goes to z0 times j omega times r/c. j omega r/c times z0. And you since z0 is rho 0 c, you could write this j omega rho 0 r.

It's totally imaginary. This z is a ratio of pressure to velocity, and in that analogy, this thing is an inductor, which we know as a mass for acoustics. So the thing behaves like a mass.

Now, if it behaves like a mass, remember what the open tube did. We were able to compute the impedance for a short, open tube. And we were then able to go back and compute it another way, as if this was a block of air in here that had no compression. And it just moved like that. And we got the same answer.

So when we say something behaves like a mass, it represents a block of air that's just going back and forth. And that's what happens when you're looking from a surface here. It's divergent. Namely, there's air moving in and moving out, moving in and moving out. And you've got a move that darn piston, or whatever is driving this thing, to shuffle all that air back and forth. And as we're going to see in a few minutes, that's the dominant motion that you have on the piston. Moving it back and forth.

Well, it turns out, if what you move the piston back and forth with, you put some electrical energy in over here, and if this process from here to the motion is inefficient-- as it's going to be spectacularly so in normal transducers-- then you're putting a huge amount of energy in here just to move that thing back and forth, that mass back and forth, and very little energy is getting radiated.

So there are shocking figures-- I mean, for example, as we'll see when we get the loudspeaker model-- I shouldn't give this away, but it'll be as spectacular when we get to it. You have 100 watt amplifier in your hi-fi system. About 99 watts of that thing go to heating the voice coil, which is exactly what you don't want. Because if it heats the heat, adhesives may age, and the coil may unwind, and you've got a mess, and no sound. So you've bought this big amplifier to heat your room and the loudspeaker, and 1% of it gets out the other end.

So you don't like to see things like that, with mass on the end. But that's exactly what's happening. You know if the impedance looks like a mass, there is going to be air just flopping back and forth here that has nothing to do with radiation.

So let's take a look at this expression a little closer. Let me-- this is a specific omega. It's on the omega axis. So I put that in form of an s-- j omega equals s-- then I can explore this anywhere in the complex plane for a moment. So let me write this as an expression. z of r and-- now I'm going to specialize for the moment. r0. Let's talk about that r equals r0. Let's talk about the surface of this sphere, whatever r it is. Let r equal to r0. r0 in s equals z0 from right here. s over s, plus c over r. Let me make a little-- this is the s-plane. Where are the poles and zeroes of this function, quick? Quick!

SPEAKER 1: [INAUDIBLE].

DR. AMAR G. BOSE: Pole. Great. OK. So the magnitude of something like this is, if I'm at this point on the omega axis here, I get the magnitude by the magnitude of this vector divided by the magnitude of this vector. Pole zero patterns never tell you the constant that's out front. It's normalized on that. So I get something that-- the s becomes, when the s is equal to j omega, I get this vector divided by the pole vector from j omega to this point. So I can right away see that the magnitude of the impedance starts out at 0, and as I get way, way up here, the vector from the 0 to omega and the vector from the pole to omega become equal, so it gets to 1 or z0.

So this thing comes up in magnitude as frequency goes along-- this is magnitude of z. I can now, if you want, call this radiation impedance because we specialize to r0, so you might want to just remind yourself that that's also called radiation impedance by putting an r on there. So that marches along here to z0, the radiation impedance, as we go along.

So very small impedance at the origin. Remember now, we made the assumption when we looked at the open tube that p was equal to 0 outside here, and I told you that it wasn't quite 0, but it's very small compared to z0 inside. And of course, this whole thing, when omega is small enough, is very small-- this impedance here is z0 multiplied by something that's very small. And so if it's very small compared to z0, the standing waves in here, gamma is almost 1, the standing waves in here don't know the difference between being p equals 0 and p equals some tiny, tiny, tiny little quantity. That's the justification for what we did earlier when we said an open tube has a 0 pressure outside, which you know, of course, couldn't be true if you had any kind of pressure in here. And you know that an open tube distending wave with a given frequency, the pressure is 0-- we made the assumption 0 here-- rises slowly, and the velocity is the constant that goes down.

We'll come back to this expression, but let's first look at the real and imaginary parts of the impedance as a function of omega. Impedance as a function of omega is over here.

That's amazing?

OK. Now, z of z radiation of r and omega is equal to z0, j omega, j/j omega, plus c/r0. Let's get that in terms of its real and imaginary parts, and you know how to do that with complex numbers now. You'd multiply top and bottom by the conjugate of downstairs, in which case you get omega squared plus c/r0 quantity squared. And upstairs you would get conjugate, that would be minus j omega times j omega, that is minus j omega times j omega is plus omega squared for this times that plus j omega c over r0.

Now it's easy to get the real part, then the imaginary part. Real part is just this, and the imaginary part is this. I could write this as a sum of this plus this. So taking the real and the imaginary parts of an impedance like this gives me the equivalent circuit of two things in series. Sum of impedance is that. So if I wanted to make a model of this, I'd have something which was real represented by that, plus, in series with it, something that was imaginary, modeled by that.

So let's just look at these two terms. If I did this on a linear scale, the real part would start off as omega squared. And if we go to a constant when omega got much, much greater then c/r0, it goes to 1 or z0. I'll just normalize it and say, I don't care what the z0 is. So I'll just say it goes to 1.

If you want, this is the z. Real part of z of r over z0 goes over to 1. Now, the imaginary part starts-- that's this part over here in the red-- that imaginary parts starts as omega, not as omega squared. And as you go to high frequencies, it goes to 0 as 1/omega. So if you actually plot it out, you'll get something that looks like this. And this would be imaginary part of zr over z0.

So at very low frequencies, what happens is the whole impedance that you see-- because these two are in series. In other words, this model here, when I wrote the sum of, I can now write zr of r0 in omega, and say that that's looking into a box here, which is real part plus imaginary part. Because the sum of impedances mean that they're in series.

So it says the real part is negligible compared to the imaginary part. As you go down in frequency for a given size ball, the world books imaginary. And that's exactly what we saw before, because as you go down in frequency relative to the wavelength, that ball looks very small. It has an enormous amount of curvature, has enormous amount of divergence, and so the world looks imaginary to it.

And it's awful hard to radiate power into an imaginary world. You can't do it. OK? If your impedance is all imaginary, try very hard. Put all the voltage you want across it. No power. Instantaneous, yes. It all comes back a little bit later.

So no average power, which is the radiated power. When you're talking about radiation, the average power that flows into a radiation impedance is the power that's radiated. Normally up until now, you would say the average power flowing into a network, into a box, is the power dissipated. But when we looked at energy, the real consideration only was for average power that it didn't come back. When we looked at power going through a line, if it didn't return at any point in that cycle, we called it dissipated, because that's all we knew at that time. It was getting absorbed somewhere. But average power, when you're looking at a radiation situation, is the power you want. That's the communication signal.

If you looked at these things, like you'll quite often see them in textbooks, you'll see this thing on a logarithmic scale. And of course, this thing goes in as omega squared, so the real part goes down as omega squared. So that's 6 dB per octave and it goes like that. And then the imaginary part goes as omega, which is-- sorry. This is 12 dB per octave, because it's 1/omega squared. And then this constant, this is 6 dB per octave, and it will go down, also, at 6 dB per octave imaginary part-- where is it? Imaginary part. It goes as 1/omega. So 6 and 6. This is the way you would see the radiation impedances listed.

Now, interesting. Very interesting, when I first came across it, and saw what the results were. I was amazed. That is for a sphere. Very neat, very clean. No approximations. You can calculate it. If you actually went back and took the piston in an infinite plane, what you find out-- ah, there's a little bit of change in the number scale here, not too much, for a radius similar-- the two sides are similar.

But what you basically find out is that it looks just like this, except here, after you cross over here, there are-- this, by the way, is about c/r, somewhere in here. Very close to this point for omega. You find that there's some ripples in here that die out like this, and there's corresponding ripples in the real part that die out like that.

But the shapes are very, very similar. They both go down like this together. You can take a piston in a tube, in a very long tube, and look at its radiation impedance, it'll look like this. Take a piston [UNINTELLIGIBLE], it'll look like this. And of course, this sphere does look like that.

And you can see these. I want you, in this case, to look, because it's a good summary, at Beranek's chapter five. He has the graphs for these. Just look at him and think about them a little bit.

There is a problem in the book, as I remember. The graphs for a piston in an infinite baffle and for a piston in a tube are reversed. In other words, the legends are wrong. So you just have to reverse them. But they're so similar in their appearance that you would never know this, unless you happened to bump into a situation where you had to calculate it, and you wanted to check your calculations with what was in the book. Then you'd find it. Otherwise you wouldn't know it. So it is interesting that these more complicated situations look similar.

Now, let's take a look at the world of radiation impedances in terms of sort of representing the world as a box with two terminals on it, and see what we would put in there. In other words, the world now is pressured, let's say, the cross-variable velocity is a through variable, and inside this box is the radiation impedance that represents the rest of the world. To find out what is inside of that box, we only need to go back to the radiation impedance that we calculated and make believe that all of that is contained in a box, and look at the impedance, and see what would be in there.

So we already computed the radiation impedance. What would you like to put in the box? And why?

Looking at the top of this section right here-- there's the radiation impedance. Now, you cannot answer the question I just gave you without asking at least another question. What's your analogy? Are you going from one discipline to another? I'm now going over, if I didn't say it, to electrical, where I have to make an analogy. So the analogy that I want at the moment is voltage analogous to pressure, which means, of course, current analogous to velocity.

So this is the impedance of a network, where impedance is defined as voltage over current. What's the network?

SPEAKER 2: [INAUDIBLE]

DR. AMAR G. BOSE: Somebody's got to know the answer to this.

SPEAKER 3: Inductor in a series with a resistor.

DR. AMAR G. BOSE: What was that?

SPEAKER 3: Inductor in a series with a resistor.

DR. AMAR G. BOSE: Inductor a in series with a resistor. Well, let's try it, see what happens. That's the nice thing about this. You can always just think of something, and then check it, and see if it checks out or it doesn't. Everybody agree that that checks out?

SPEAKER 4: No.

SPEAKER 5: [INAUDIBLE]

DR. AMAR G. BOSE: Where doesn't it check?

SPEAKER 5: [INAUDIBLE]

DR. AMAR G. BOSE: [UNINTELLIGIBLE] say no.

SPEAKER 4: It should be parallel.

SPEAKER 6: 0 at the line.

SPEAKER 4: Parallel.

SPEAKER 7: Parallel to [INAUDIBLE]

SPEAKER 8: Parallel to [UNINTELLIGIBLE].

SPEAKER 9: [INAUDIBLE].

DR. AMAR G. BOSE: Ah! Zero at omega 0. This thing has 0 [UNINTELLIGIBLE] at omega 0? No way. So that can't be it. Now how does 0 [? repeating to the ?] omega 0? What do we have to have besides the short circuit?

SPEAKER 10: [INAUDIBLE]

DR. AMAR G. BOSE: Not if they're in parallel. All right. What else?

SPEAKER 11: [INAUDIBLE]

DR. AMAR G. BOSE: Will this do? It's hard [UNINTELLIGIBLE] in parallel. Now, two elements in parallel, any two different kind of elements, be they an l and a c, and l and an r, a c and an r-- they have a pole at the point where two impedances are equal and opposite. So those two impedances are equal and opposite when s is equal to minus c/r. That's the pole. The pole means that the whole thing goes infinite, the denominator goes 0, OK? So that happened at s equal minus c/r, as you can see from here. Here's the pole. That's how we got it in the first place.

So this thing has a pole, then, at that frequency. It has a 0 that's obvious at the origin. Let's see if we can figure out what the elements must be. What's this element?

z naught. Sure. Why? Because when you go to high enough frequencies, you look at this network here, and you go to high enough frequency, this increases, becomes millions of times larger than this one. All of the [UNINTELLIGIBLE] goes down here, so this [? develops ?] a resistance. He can't tell the heat's here. He doesn't draw anymore [UNINTELLIGIBLE].

So looking at the network, you know that this network goes to this value, whatever the r is at high frequencies. And you know that we figured out before that the radiation impedance at high frequencies gets very, very large, goes to z0. So the resistor then, which this network goes to at high frequencies, must be z0.

OK. And so the inductor now? Well, if you have one, you know what the pole frequency is. You can all figure out what the conductor is. You can figure it out from the pole frequency. You can figure it out another way, if you wish. You can go to very low frequencies, and at very, very low frequencies, this thing is out of the system. He's very small compared to this. And so the whole impedance goes to j omega r0 over c. j omega r0 over c times z0. z0, j omega, r0 over c, or j omega-- if I put in for z0 rho 0 c, j omega rho 0 r0. So this fellow here, then, can be expressed as rho 0 r0. And that's the network that matches that radiation impedance?

Any questions up to now? Because this is one trivial, almost, example algebraically, but it has the heart of radiation impedance in it. And if we understand just the things we've talked about here, we will have a very good feeling, and not have to be afraid of radiation impedances in any field. It's just a matter of how complex it is to calculate them. And I'll talk about calculating them and making models of them in a minute. But here's one in which we were fortunate, in respect to A, it's simple to calculate, and B, which isn't obvious to you yet, it has a model that's exact with resistors and, in this case, inductors.

Or if you want it, by the way, suppose I want a model now of this thing-- and we'll need that-- in which velocity is the across variable, and pressure is the through variable. Would you please give me the model? This is the box.

SPEAKER 12: [INAUDIBLE]

DR. AMAR G. BOSE: z0. Value?

SPEAKER 12: [INAUDIBLE]

DR. AMAR G. BOSE: [UNINTELLIGIBLE]. Now, turns out, why would you need this? Well, you'll need this because the most convenient models-- for example, if you want to talk about what drives that piston over there, you have an electromagnetic transducer driving it, the natural, and we'll see we're forced into it, across variable then that will be-- well, let me not say we're forced into it. The natural one will be one with velocity as the across variable. And you want to get the whole circuit together, acoustics and mechanical and electrical, and you need a circuit of this form to hook it on the end. There you are. Thus the use of the dual networks that we were talking about.

Now I just told you I was going to do something. Oh, yeah. Let me first-- because it's just so important, at least I feel it is, review what we've done here, and then let me extend this to the much more interesting and general cases. What we did here is we had a sphere, we computed-- in essence, I computed the impedance out here. But what I was after is the ratio of pressure to velocity on the surface as a function of omega. It's now fixed at r0. Got the pressure to velocity ratio out here, which is the impedance seen by the sphere as a function of frequency. Took that function and said, oh, now imagine that it's a network. Looked at the network, constructed a model, analyzed this in terms of its real and imaginary parts so that we would understand, when we need it, why it's hard to drive something at low frequencies when there's no real part. Without a real part of the impedance, you don't radiate any power.

Now, you know-- by the way, this is just like the motor thing we talked about. The industries, you can go into a big machine shop, go into where the power comes in, and find a whole bank of capacitors sitting there, because they want to have a unity power factor for their bill. And they would put capacitors across here. You could tune such a device with another acoustical device, it turns out. I'll let you think about that. How you might take a piston in an infinite baffle, for example-- what could you build out here that could tune this thing so that that piston saw a real impedance? It's a very interesting problem. I don't think you'll find it in textbooks or anything. It's not difficult. It's within the realm of things you have done in the class.

What structure could you build out here such that at a given frequency-- you pick the frequency-- that this piston would see a real world? And that would be very interesting. Because I told you, the efficiency will go way up then. Now, since it's only at one frequency, I hope you're in the business of building sirens or tone generators, because that's all you're going to get. But it is a very nice problem. It will exercise your creativity. Find a structure that would do to your impedance what the machine shops do to their impedance so that they see a real one and have to pay less.

For example, do you ever think of why a digital watch makes that unpleasant ring? It could do other things. Could even talk to you. Could tell you the time. You could make an extremely effective resonator, highly efficient, when you want only broadcast one frequency. The minute you want bandwidth, totally different thing in terms of power. So your watch makes just a beep. It models an LC circuit. High QLC circuit. Doesn't need much power to drive it when there isn't much in there.

So whenever you have a job to design something that's very narrow in frequency, you can do it with efficiency. Which if it was broadband, you'd have another problem.

OK. Now I keep going on the sidetracks, and I forgot what the heck I want to do.

Want to talk first about a little peculiarity, if you wish, of-- I hope I didn't erase it. Let's see what's under here.

Well, the pressure is a peculiarity of this thing. It doesn't lose us any generality in our thinking, but the sphere that we analyzed has a situation that you don't encounter otherwise. Namely, it has no near field, if you want to say it, in pressure. The complex amplitude as you go out in pressure just get smaller, that's all. So in time, if you look at that, if you radiated a wave that was a triangle here, and you neglect the fact that this end of the triangle was a little further way than the other end, that's a small thing compared to the big R-- that triangle just goes out, and it gets smaller and smaller.

So there's really no near field in pressure in a spherical wave. If you did the analysis for any of the other situations, you would find that there was a near field in pressure, also. It only has it in velocity. If you look at the expression for velocity, take the j omega-- where the heck is it? I have it upside down. But if I flip this thing upside down, and flip this upside down, I would just solve this thing for u of omega. Take this over here, bring this over here, and flip it around. You see the view of r in omega is a big function of omega. And it has this stuff in it, which when you get the large omega, it becomes real. But for small omega, it's a mess. It's real and imaginary.

That's the near field. I mean, you're out of the near field in antennas, or in acoustics, when the impedance seen looking out there is real. When you get to such big wave fronts that the thing looks real.

So if you just write the expression, solve this, just bring this up, bring this over here, and turn it upside down, you notice that the u has the near field in it. And the near field in the u is caused by this stuff. This all becomes real when omega gets large enough.

And in general when you look at impedances, looking out, radiation impedances, you'll get the terms that are higher order in this, even. In other words, you may have an omega/c plus omega/c squared, cubed, whatnot. For oddball shapes or real--or not oddball but real shapes, but not so simple mathematically. And those are all associated with the near field, and they will be, in general, in both pressure and velocity. There's no complication in our thought process, but it's something that I don't want you to think just exists always in the velocity, and not in the pressure. Or in the e field and not in the h field.

So now what you do in general is, you have to find, from whatever source you have, the total ratio of whatever you want, the force, the velocity on this oddball shape, or whatever it is. When you find that, you'll get a function of omega, but you won't be so lucky that you get a nice rational function, in general. You'll get, God knows, depending on the geometry, some function.

And what you do to get even the models that are shown in Beranek for the piston and the piston in a tube, piston in an infinite baffle, what you do, then, is you take this frequency function, which has a magnitude and a phase, as you could get from it, and you make a rational function approximation to that given frequency function. And there are various ways you can do that. You have to sign an error criterion. You might take a mean square, or you might take a weighted one. But you make an approximation.

And what you would normally do is, perhaps you'd start with the magnitude of this. What you found. In other words, you found a very complicated, strange function. And you say, well, I will allow myself, let's say, five poles and maybe five zeros, or three zeros, or whatever. And with those poles and zeros, you would s minus s1, s minus s2, s minus s5, let's say. And upstairs, you might have s minus sa. s minus sd, or whatever.

And what you would do then is you would select the a through d and the 1 through 5 to give you the best, by your criterion, approximation to the actual magnitude function, whatever you had to, over some frequency band. And once you make this rational function approximation, then, of course, you have a network. A network which all of you can synthesize, maybe not as one big network, but if you want to get very simple, you can take off this part, and you can synthesize that. Then you can cascade it with the current source, an op-amp or whatever, with the next set of them, and the next set of them, et cetera.

But you can realize that as a lumped parameter network. You actually can realize-- I mean, it can be shown that you could realize this without any active things, as well. Just as rs, ls, and cs.

And so you then take this thing, and gives you some sort of a complicated network. You'll see minimally complicated ones-- here's one that you'll see in Beranek for a piston and an infinite baffle. You'll see something like. I'll draw one for which velocity is across. The reason that you know it's velocity right away is because I put a capacitor in there. And the mass, which all of these impedances are going to see, is an inductor in the analogy that velocity is the through variable. So if I put a capacitor in there, you know right away it's going to be the across variable's velocity and through variable's pressure or force, whatever you want to make it.

And you see a model that looks something like this, for example. Let's see. I have to be a little bit careful. To tell the truth, I'm not sure what you see. You have to look it up. But this might be one that you'll see.

If this were a model that I saw for some radiation impedance looking out from some device, what resistor in this thing could you tell me the value of right now?

SPEAKER 13: [INAUDIBLE PHRASE].

DR. AMAR G. BOSE: Parallel combination of this is z0, because this thing becomes a short circuit as you go to high frequencies. The impedance will get real as you go to high frequencies. And so the impedance of this at high frequencies is this in parallel with that.

If I had a different network now-- let me just look at this. I'll try to make up one here. This is an impedance now. Since there's an inductor here, the through variable must be u. The cross variable is some kind of pressure or force. So what can I say about the elements in here at high frequencies?

Any element. Somebody else this time?

SPEAKER 14: [INAUDIBLE]

DR. AMAR G. BOSE: Yeah. Now, the other thing. If you compute-- let's say you have a loudspeaker now, and you've analyzed the thing, you've found out what the velocity of the piston, of the cone is. And somebody asks you, well, how many watts acoustical power is being radiated by that thing? Well, if you have a model, all you have to do, is the radiated power from the loudspeaker must be the power that goes into this box.

Because average power flowing out of these terminals in terms of the transducer is radiated power. Average power flowing past this border in terms of the model is power dissipated in all the resistors. So at any frequency, if this was a decent model for whatever frequency you're dealing with, to find the radiated power, if you knew the velocity, all you have to do is say, well, there's u amperes going through here at that given frequency. Compute the power in here and the power in here, and that's the radiated power.

So it's very nice to have a model like this to think in terms of. You can see, for example, what happens at low frequencies versus what happens at high frequencies. You can see how the power changes just by, again, using your insight from electrical networks to tell you now about things like radiated power.

So there's a very nice reason to look at the rest of the world like models, like electrical models, if you're an electrical engineer. Again, it's taking a discipline that you got way back in network theory, and showing you can use it for problems that you wouldn't have imagined that it would be applicable. Questions?

I'm just trying to think whether I can launch onto another thing or not. I basically said, I believe, all that I want to say about radiation.

Let me tell you, I don't know whether I'm going to even bother to do directivity factor in calculations. I don't know. I'll think about it before next time. But I'm calculating something that you're never going to need, and that if you ever did need, if you understand what it's all about, you could get it like that. So I'll just discuss now what directivity factors, or directivity-- there's two words for it. Directivity factor, and there's directivity-- one of them's a logarithm of the other. I don't know what they call them. But it's a measure of directivity.