Amar G. Bose: 6.312 Lecture 08

SPEAKER 1: Quizzes that helped me a lot may help you. At the time I went through here, you used to have to take five subjects a term, and some of us in the sixth year of doing that had realized we'd seen a heck of a lot of quizzes. And so we somehow together developed an idea for preparing for them. And there were about four or five of us in the group. The idea was that we would each independently make up three problems, as if we were making a quiz to be given out. Then we would meet the night before, and we would each look at and work the problems that the group had originated.

And it turned out that this was successful beyond any of our imaginations, so successful that-- this was my sixth year, obviously. I was in graduate courses. But it was so successful that during the quizzes, we would be holding up fingers saying, we got two out of three. And I can remember quizzes where we got three out of three. I mean we just had it cold. And the basic philosophy was when you sat down to make up a quiz, what you have to do, if you imagined yourself as the instructor, you'd go, what are the basic concepts that we've covered? How could I best illustrate those concepts in a problem?

And that's exactly what the teacher does, as I found out later. He just sits down, looks at the list of concepts, and says, look, maybe we can't get them all in, but let's try. When we sit together to do this, we'll take each of the concepts and we'll assign two to one person, two to another, two to me. And then we all come in and we look at all the problems together, and then we select down from that.

I mean, I just wish that we had discovered this many years before, after it worked so well. But when you can get three out of three problems, it was amazing. And the reason we held up the fingers was, ah, we beat the system. We really got it all down before we came into the quiz. So it might be a very useful thing for you also.

Another topic that has come up, apparently, in some of the sections is, why don't we print problem solutions? We don't do it for a very definite reason. When they're there, there's always a tendency, if you're pressed in time, to say, oh, well, I'll go look at the solution. And then what happens is you think you are understanding the principles, but following and generating are two different things. So when you get out later and start working, there aren't going to be any solutions. You have to generate them. And you can't follow them, or anything else.

So we have teaching assistant time that you can talk about anything. You can come in, ask questions afterwards about the problem. Some of you ask them before. That's not a disaster either. Basically we want you to generate that. We don't want to put up a crutch in which, if you're pressed in time, you think, well, that's OK, I can understand it by reading it. You do not get the same understanding.

In fact, what you'll find out is what all the teaching assistants have just been telling me this year, and my own experience also, that everything you do as a student here, including your preparation for doctorate exams, won't begin to give you the understanding that you will get if you go into teaching. And in that respect, though I studiously avoided all teaching assistantships because I didn't want to be a teacher-- but if I had to do over again, I wouldn't do that.

If you become a teaching assistant at some point, and not just grading problems but meeting with students, you're going to find out that so much of what you accepted was just that-- you accepted it, and you didn't really understand it. And now when you're faced with explaining it to somebody, you realize that for the first time. When I came into the classroom in that first year, my level of understanding just took a totally different level and meaning to me. So that is one of the things, if you have that opportunity when you're in graduate school, if you have the opportunity where you can do it with meeting the students, not problem creating, take advantage of it, because it'll help you in everything you do. It defines a new level of understanding for you.

OK. Let's see. We ended last time with this expression for an array of sources, which was just two point sources, some distance out. Way out here, you were measuring the pressure. This was r to the center. And this was a total separation of d and a total separation of phi between the phase angles of the point sources. In other words, Vv1, Vv2. Vv1 was equal to Vv magnitude. Both of them had the same magnitude, e to the-- I think we did it this way. Plus j phi over 2. And Vv2 was the same magnitude as Vv1, e to the minus j phi over 2.

I told you that when you make up a raise, like in electromagnetics, you make a whole bunch of dipoles like this, today, and you steer them by just altering the phase. You always make these magnitudes the same. You're putting the same amount of power into each one. It's simple to do. The reason that you didn't do this, steer them electronically before, is that it was very difficult to make amplifiers. Today you can make reasonable power amplifiers in very small size, and so you can, instead of of having one amplifier feeding a bunch of antennas, all in phase, in the whole array, you can now make ones that have the same amount of power, but one phase here, one phase here, one phase here, one phase here, and you can steer the beam all electrically.

Now when we speak of steering, we didn't actually get there yet. We'd just about arrived at the end of the last time. This is the theta dependence right here. Obviously, there's no phi dependence. It's symmetrical around here. So this is the thing that we're interested in looking at, to see what we call the pattern. And you can see that that pattern will depend upon the phase angle between the two point sources. This function of theta is dependent upon the phase angle. And that's what we want to look at, and that's what generates the so-called polar pattern.

Now a polar pattern. What does it mean? It means a pattern whose-- let's just draw one arbitrary one on here first. If this is a polar pattern of electromagnetic antenna, or a light beam or an acoustic wave, what this means is that the strength of the variable that you're plotting-- it might be e-field a mile around an antenna in a constant circle. Here's a point source in acoustics. You go a good distance from the point source, a number of wavelengths at least. You go around the thing in a constant-radius circle. But the radius in this polar pattern diagram, to the intersection point of the diagram, is a measure of the strength of the variable that you measured when you walked around the large circle.

So this radius and the circle and the radius at which you take the measurements have nothing to do with each other. That's a constant. So all that this says is, on this axis, this is bigger than this. So by that much, the signal measured along this axis is stronger than the signal measured along this axis. The signal measured along this axis is zero. So origin to intersection point, that is called a polar plot of the source.

So that's what we're interested in up here. All the rest of the things just-- well, take a good look at this. If you fix a direction, theta, and you fix your phi and you have a given power into this thing, or a given volume velocity, everything that's out here, fixed frequency, is a constant. And this is exactly the expression for an outgoing spherical wave.

Now you say, well, how can that be a spherical wave? Well when you get far enough out from any of these sources-- let's talk acoustics at the moment. You're having motion of the particles and a pressure associated with that. When you're far enough out, these particles think that the other ones are behaving just like a sphere, like they would on a sphere. And what these do and the impedance that this sees is a function of what the neighbors are doing. The impedance that I see here with my hand when I move my hand forward-- it makes a difference if the other hand is going with it or whether I'm doing that. That affects the impedance here, if I'm going like that, as opposed to going like that.

So when you get far enough out, you have to go such a distance to get any kind of a theta change in here that these particles that are here think the others-- as far as they know, they're doing the same thing. And it's like a sphere radiating outward. And that's why you see expressions like this . But these are only good, you remember, when r was much, much greater, as we showed last time, than d, so much greater that all these lines are essentially parallel, and theta was here, and theta also was the angle to the point p from the center of the array.

OK. So just to see how this looks in a couple of cases, let's just take two examples. Example one, let's take d equals lambda over 4 and phi equals pi over 2. Phi is pi over 2. That means that this fellow here is 90 degrees in phase. This is a given frequency, omega. They're both sinusoidal, and this one is just leading the expansion of the other one by 90 degrees, pi over 2. And they are a total distance across from one to the other, lambda over r, a quarter wavelength.

OK, formally, we plug in. Let's see what happens. The whole polar pattern can be obtained just from this expression. The rest we don't have to worry about. It's not a function of theta. OK, lambda over 4 for d in there. Lambdas go out. By the way, just make sure that that's the right expression, because I remembered that from last time. If it's not right, I'm going to get funny polar patterns. So we have cosine pi over 4 when d is lambda over 4. Sine theta minus phi over 2, which is pi over 4.

OK. Well, let's take a look at-- let's take theta equals 0. Let's do it this way. This is the angle theta here that I'm going to draw the polar pattern for. Theta equals 0. Sine theta is 0. Cosine of minus pi over 4 is 0.707.

Let's take theta equals minus pi over 2. So sine theta is this over this. So sine theta should be minus 1. So minus 1 here is minus pi over 4, and minus pi over 4 is minus pi over 2, cosine is 0. And if I do it for theta equal to 180 degrees, I will get 1 over square root of 2. If I do it for theta is equal to pi over 2, sine theta is 1. And so that gives pi over 4 minus pi over 4, which is 1. So I get a pattern that looks like this. You can fill in the other points. That, as you have seen before, is known as a cardioid pattern.

So what happens is this antenna array up here, for these conditions, radiates strongly in the vertical direction, upward. Nothing in the downward direction. Now by the way, if you were really to go out and measure this thing, what you'd find out is this. It wouldn't be 0. And remember the approximations we said, when we pull this r out from underneath-- this is like that integral which we did for the whole body. We said, watch out, if you're going to say out there that if all the r's for all the different particles on this thing are equal, then when this particle and this one over here are going opposite to each other, you'll get a 0.

But really it shouldn't be a 0, because look at this. When this one is opposite of this, this one's going plus, this one's going minus, the way we derived everything, you get 0 out here, but clearly you get a positive number out here because the distance to this one is less than the distance to that one. They don't exactly cancel. So that's the kind of difference you get from this expression in the real situation, only because of the approximation you made. And that's a very small thing in practical case.

OK, let's take one other-- oh, by the way. You can look at these things and check that you're getting the right result. This was 90 degrees ahead of this one in phase. Phi over-- no, no, let's be careful. Is this the sign that I used? I just want to make sure. It's a plus sign here and a negative sign here? OK. So this fellow is 90 degrees ahead of this one. OK, now when the wave travels from here to here, this one loses 90 degrees. Are you sure these signs are right? I tend to get 0 here and maximum down here.

SPEAKER 2: [INAUDIBLE]

SPEAKER 1: That's it. OK. Check your notes. Was that how we derived it? OK. Good. So he starts off behind by 90 degrees of this one. By the time this one gets up here, he loses 90 degrees, so they're in phase. So you have a maximum going up here. Now watch the other way. When you go down here, this one was already 90 degrees behind this. By the time he reaches here, he's 180 degrees behind this. So how much is going down here? 0. And when you're out here on this axis, both of them are going the same distance, but this one and this one are 90 degrees apart. And so it's only the square root of 2 that you have on the axis. Whereas when they went together in the up direction, you got 2, and you only got the square root of 2 when you went out. So that's how you get this kind of a polar pattern.

Now here's one you could almost do from inspection. Let's just take one more example. d is equal to lambda over 2. And phi is equal to 0. d is equal to lambda over 2. They're a half-wave apart and they're in phase. So now when I go up this way, this one loses 180 degrees, because it was a half-wavelength here, by the time it gets here. So it's exactly out of phase with this one, and there should be no radiation upward. If we go downward, again, they're in phase to start with. By the time it gets down there, it's 180 degrees. They're out of phase, so no radiation this way. If I go out here on that axis, they're in phase, so they add. I get 2 that way. I get nothing down or up. If I go out this way, I get maximum. And so if you-- figure-eight antenna. The polar pattern looks like that. No radiation this way, no radiation that way.

So you see, just with two elements-- and I've only given two examples of those two elements-- look how dramatically you can change the polar pattern. And this is actually used in some cases, because there are radio stations which cater to traffic from one direction at one time of the day and another direction at another time of the day. All you have to do is change the phase of your array of antennas. You've seen out there maybe on 128 AM, the antennas are spread very far apart because the wavelength is huge, 300 meters per megahertz. And if you change the phase between those things, you can alter the polar pattern any way you want.

And think about it now. It's very important, because if you have a city like Boston, here's the ocean here. And you're broadcasting here. And the boonies are out here. You don't want to talk to them. remaining The main public is here, so you want all your energy directed into the city. So putting a given amount of power into the antenna is extremely important to direct it where you want.

And the same thing happens in acoustics too. You may have alarm systems that-- yeah?

SPEAKER 3: [INAUDIBLE]

SPEAKER 1: Yeah, there's conservation of energy, but energy, as we will see, is the total power radiated out of the thing, if you put these arrays and you fool around with phase of them, insofar as the one doesn't affect the radiation impedance of the other. Total power is the same. Now, power turns out to be the integral of p squared over the polar pattern. So it looks like that, but it's not quite area.

OK. That's as much as I really want to say about arrays. You can extend it. Obviously if you get a huge array with many, many, many-- the problem becomes the same as what you've seen in physics with diffraction, a diffraction grating. You remember that it all comes through the holes there, light. And you have the d sine theta business. And now the thing is, you have so many of them that what you get is a pattern that looks like this. If you change the angle, it'll look-- and then just a little, like, sine x over x. The more you get, these ripples get very fine. And so with a diffraction grating, of course, you can take a given frequency and put it-- as you've seen, when you look at light through those things, the given frequency will go right here. The next frequency will go here, the next frequency goes here. You do the same thing with antennas, with enough elements.

OK. Now let's talk about energy. This turns out to be much more useful than just a computation. Many problems in all the different disciplines-- well, there are a subset of problems that yield themselves to energy considerations that, really, it's the easiest way to look at them. You've already seen that some problems it's easier to look at the frequency domain than time domain, and maybe vice versa. Energy not only gives you a basic guide to what's going on, but sometimes it forms an easy way of looking at things. So let's talk about the energy in sound waves.

Go back to our little particle, this imaginary surface that we have surrounding a very, very small amount of mass. A very, very small volume, but it follows a constant mass. So if I wanted the energy density per unit volume, if I take a look at that little particle, I have kinetic energy and I have potential energy, because if I squeeze the volume, that's potential. If I move it, that's kinetic. So the D, which we'll use for energy density-- i.e. energy per unit volume-- and we'll write the KE one. That one turns out to be very easy, from what you know just from physics. 1/2 mass, and the mass in this thing-- mass per unit volume, because this is an energy density, mass per unit volume, of the particle, a little v0, times the velocity squared. I'm using velocity here as a scalar. Simply, I'm doing one-dimension waves.

OK. So this is 1/2 rho 0 u squared. Kinetic energy per unit volume, if I have particles moving. Potential energy per unit volume, DPE, is-- now, let's see if you remember. Potential energy-- if I have a pressure on a balloon, and I squeeze that balloon down from outside, how much potential energy do I get? Do you know this? How much potential energy gets stored in this thing, in this balloon, if I come along and I squeeze it? In terms of a volume change, if I just squeeze it a little bit? d tau?

OK. I'll give you an assignment. If you don't remember this, go back to one dimension. This is the x dimension along here. Put a little piston here, and figure out-- the potential energy increases the work you do when you push this thing, Dx. So there's a certain force, a certain pressure, that's inside here, that I push it just a little, Dx, and I find out how much work I do. That's how much potential energy I put into this thing.

So you do that, and then I want you to come from that to integral of pd tau, where tau is the change in the volume of a general form.

Now this is a minus sign, because if I squeeze the volume, the d tau is negative. So if I did it without the minus sign, pd tau would be the work that this fellow did on the rest of the world. But I don't want that. I want the energy that's going in, so it's pressure times the negative of d tau. d tau is an increase in volume.

OK. Now that's unfortunate, because it's not quite so easy as this, which just fell out on our lap here. pd tau. But how do I get all this stuff? How do I look at this integral? I have to get this into some kind of a situation that I can do something with it. I have P and I have tau here. What do I do? Tau is volume, right? Any other relation you know between pressure and volume? Tesla. P total-- here we go again. Pv total to the gamma is a constant. So I have to use that, and I can eventually get rid of this d tau and put it into pressure. Let's do it. I want to get out of d tau eventually, so let's just take the differential of this.

Let's see now. dPT-- remember, PT, the total pressure, from the very beginning was P0 plus the pressure due to the sound waves? So dPT is going to be dP. But dPT times VT to the gamma plus gamma VT to the gamma minus 1 PT times dVT. But dVT is the d tau, because VT was equal to V0 plus this crazy thing, tau. So I remember that dPT is dP, take this, and just get an expression for d tau.

So d tau is equal to minus-- I'll bring this over, take it over to the other side. I get minus VT over gamma PT times dP, which is equal to dP tau. VT, up top-- this is just one term, not a sum of terms, so I can approximate that by V0. Similarly, downstairs, multiplying everything, I can use-- there's something like 10 to the fifth or 10 to the sixth difference between the P that's incremental [INAUDIBLE] a sound wave, and the P0. So I can write this as d tau is equal to minus V0 over gamma P0 dP.

OK. And now put that in here, and we get DPE-- aha. I said I was going to calculate the potential energy, DPE. density-- in other words, potential energy density, just like kinetic energy was 1/2 mu squared, kinetic energy density is 1/2 mu squared over V0, so this has to be over a V0 also. In other words, potential energy per unit volume is the definition of DPE.

So let's put this in. All that we have to do is substitute for d tau this quantity. That's a minus sign. There's a minus sign out here. So I get-- there's a V0 upstairs. That'll cancel the V0 downstairs. So I get Pd tau is going to be P over gamma P0 d tau. Gamma P0 comes out front, and so this is of the form, then, 1 over gamma P0, 1/2 P squared plus a constant, indefinite integral. That's the answer. Hold it. Just let me put it in the integral form. I already did it, but 1 over integral P squared, dP is equal to 1/2 over gamma P0 times P squared plus a constant. And the constant is 0, because we are defining the potential energy that results from the presence of the sound wave. When there is no P, there is, by definition, no potential energy in this. We're only talking about the potential energy that's due to the presence of the sound wave.

SPEAKER 4: [INAUDIBLE]

SPEAKER 1: Oh, yeah. I put down the answer first, you're right. OK?

SPEAKER 4: [INAUDIBLE]

SPEAKER 1: This one should be-- oh. Thank you. This is integral Pd-- let's start over. God. OK, we want to start from here. Let me copy it all down and not try to do too many steps. Pd tau over V0. OK. Now I substitute for d tau. V0 over gamma P0. And that's a minus sign and dP. So that gives P. The V0's go out. Gamma P0. Pd tau-- I'm only substituting this-- gamma P0 dP. And the integral of that is equal to 1/2 over gamma P0 P squared, plus a constant. And the constant, as we just argued, is 0. Pardon me. Thank you very much.

OK. Now by the way, I could also write this 1/2 over rho 0 c squared, because gamma P0 is rho 0 c squared, by the definition of c. Remember, c squared was gamma P0 over rho 0. So this could also be written, then, as this quantity times P squared for DPE.

OK. So now if we have both of these quantities-- I'll just write them down again. DKE is equal to 1/2 rho 0 u squared. DPE is equal to 1/2 over rho 0 c squared times P squared. That is for any kind of a wave, any kind of a disturbance that has a velocity associated with it and a pressure. The velocity gives the kinetic, and the pressure gives the potential.

Now this is the fundamental result. What we want to do now is just look at this in two cases, a traveling wave-- a plane wave-- and a standing wave in a tube. The expressions for both those, we've already derived. And all we need to do now is plug those expressions for pressure and velocity into here, and we'll learn something about what is the energy in traveling waves, and what is the energy in standing waves.

OK. So for plane waves that are traveling-- plane, travelling-- plus going, we have for p of x and t, P plus cosine omega t minus x over c. There can be a phase angle to it also. Let's see. j phi, if we want. And for the velocity wave, we'd have a plane wave, u of x and t. We'd have P plus over z0, which is rho zero c, cosine-- same thing-- omega t minus x over c plus some angle phi. The angle phi, in terms of the complex amplitudes, was equal to p plus magnitude e to the j phi, if you remember.

OK, these only differ by z0. Remember, they're real, because the impedance faced by a plane wave is real. So we have to stick these into here and see what happens. Let's do the DKE, this one here. I think I can do it up here. It'll save a little space. DKE is equal to 1/2 rho 0 u squared. That's P plus magnitude squared over z0 squared. I'll do that as rho 0 squared c squared. And cosine squared-- well, cosine squared is 1/2 1 plus 2 cosine, right? So 1/2 1 plus 2-- sorry, 1 plus cosine 2 times the argument. Cosine 2 out in front of the whole argument that we have in here. Omega t minus x over c plus phi.

And DPE is-- oh. Yeah. Oh, no, I did use the u for the-- that's it. OK, now the DPE is using pressure squared, with 1/2 rho zero c squared outside. 1/2 rho zero c squared downstairs. Pressure squared is just P plus magnitude squared, times the square of this. And the square that is, again, 1/2 times exactly the same argument, 1 plus cosine 2 omega t minus x over c. This looks bad, but you'll see very soon it all clears up. Namely, look, first of all, at this, what's out front here. If I cancel that rho 0 with that one, everything to here is identical. It's 1/4 p plus magnitude squared over rho 0 c squared. So we can write out these expressions, or we might be able to just look at them and see what we want.

First claim is all of this between my hands is identical. [INAUDIBLE] OK. Now what is the time average? I'll denote time average of DKE this way. Time average. What's the time average of this fellow here, in terms of what's out front? What's out front. Yeah. The time average of this thing is clearly 0. If I integrate it over time, basically, it's 0, even if I integrate it over a short amount of time-- equal to a period of course. But in the limit, this goes away. Time average of the cosine is 0. So it's what's out front.

Now what's the spatial average? In other words, if I looked at this-- here comes this wave across here. This plane wave. I take a photograph of the whole thing. I see what the pressure is, going like so, at that instant of time. What's the spatial average of this quantity, in terms of this? Same thing. You average with respect to x. I don't care what the argument is. You keep going with the argument of the inside of a cosine function, and you get a 0. It's a cosine [INAUDIBLE] So this is equal to DKE, space average.

Now since what's out front here is exactly the same for both the kinetic and the potential energy, we can write equals DPE time average equals DPE space average. So the average in time of the kinetic energy is equal to the average in space. And it's equal to that in the potential energy also. OK. That is a useful thing in the sense that when you're thinking of computing what power flow is-- power is energy per unit time. If you're looking at this point in space, a plane wave's going by here. If you know what the average energy is back here, and you know what the velocity is, you know how much of it got across the border that you're looking at per unit time, which is power, when we get to it.

It's not even worth writing down what they are. You know all of these averages are equal to 1/4 P plus magnitude squared over rho 0 c squared. Any one of those averages. So DKE plus DPE is D, the average energy density in the plane wave. Spatial averages equal time averages, because of what we have over here. And that, then, is equal to 1/2 the total energy associated with the wave. P plus is DKE plus DPE. P plus squared over rho 0 c squared.

So much for a plane wave. Now let's look at one more case, namely the standing wave. Standing wave, we did--

SPEAKER 5: [INAUDIBLE]

SPEAKER 1: Yeah, yeah, exactly. At any point. If I pin a point down in x and I look at the energy density at that point, it's going up and down with a twice frequency cosine. These are only averages. Yeah. You?

SPEAKER 6: [INAUDIBLE]

SPEAKER 1: They look exactly the same. Yeah. I don't know if there's supposed to be, but that's the way they come out. Because when you try to move the wave, you build the pressure up. And the pressure and the velocity in the plane wave are in phase. And that's what happens [INAUDIBLE] the signs are the same.

SPEAKER 7: [INAUDIBLE]

SPEAKER 1: Oh, yes. Thank you, thank you, thank you. Do anything you want. What do you want? Bar? Good. Or we can put the hat over it. It's all the same. We could, just as a reminder. OK.

Now let's do a standing wave and see what happens. I think would like to go back to the first board here to do that. Oh, you know, I was just thinking as I was erasing the board, I think I told you when we started in on these point sources that I would show you this very interesting thing about near field. I thought I was going to squeeze it into it, but I'll come back, because I really want to do that when you have energy behind you and we can look at what this near field means from more perspective then we have now. So we will return to the ball and see what impedance it looks at the world with later. I won't do it now.

OK. Standing waves. We'll take the case of the closed tube, which we did many times. And the pressure for negative x, x and t, was 2P plus magnitude cosine omega x over c, cosine omega t plus phi, where P plus was equal to P plus e to the j phi. I think that's correct. Yeah. And the velocity that we derived, x and t, was the same thing except-- well, not the same thing. 2P plus over z0, which I'll write as rho 0 c, sine omega x over c, sine omega t plus phi. Sine, because, of course, at the x equals 0, the bounding condition was velocity was 0. And we found that the pressure doubled. P plus was the incoming pressure. It smashed into the wall, and that generated another P minus going back, equal to P plus. So you got a 2 out here.

OK. Those are the expressions. And the basic DKE and DPE are there. So what we want to do now is just plug those in, and then we'll have an interpretation that is useful for standing waves. DKE-- 1/2 rho 0 u squared. OK. DKE. 1/2 rho 0-- upstairs? Yeah-- u squared. u squared is 4P plus magnitude squared over rho 0 squared c squared, times sine squared omega x over c, sine squared omega t plus phi. OK. And DPE is 1/2 over rho 0 c squared times P squared, which is 4P plus magnitude squared cosine squared omega x over c cosine squared omega t plus phi.

OK. Now these are double frequency terms. Whenever you see energy, that's going to happen, because if you have a spring of a fixed length here, and you squeeze the spring by a given Dx, you store energy in it, which can come out again. If you take the same spring and expand it by Dx, you store energy in it, which can come out again. So if a spring is moving like this, clearly the energy function is going to be a double frequency function, because it didn't matter whether the spring got compressed or expanded. The energy went in either way.

OK. What I really want to look at is to plot this thing, plot the envelope. Remember, the envelope-- this is the only part that's time function here-- the envelopes have to do with everything that's out here. And the envelopes of this are exactly the same. In other words, this and this are equal. 2 cancels out here. 2 goes in there. So these two are equal.

Now the envelopes, then, look like the following. And here, again, now, we're going to get a very interesting comparison, as you always should be able to make between waves and lump parameters. Namely, let's look at our wall here, or tube, whatever it was that we derive these for. If I were to plot the potential energy, I would have this. Cosine squared never goes negative, OK? If I were to plot the-- let's write this down. This is DPE potential energy. DPE envelope. And if I were to plot the other one for DKE, I'd get this. Same height. DKE envelope.

Now they are orthogonal in time also. Watch. These things are orthogonal in space, but they're orthogonal in time, because look, sine squared is 1 plus 2. Again, it's a double frequency term, if you just expand sine squared out. And if you expand cosine squared, it's a double frequency term. So they're orthogonal in time and space. So what actually happens here is at one instant of time, you have potential energy very high here, and then at the next instant, this potential energy goes down to 0 and kinetic energy occurs over here. So potential is being swapped in a standing wave for kinetic, and it's going like this. In other words, as the energy goes from here to here, it gets to be kinetic energy. As it goes from here to here, it's potential energy.

And this should be no shock because take just a capacitor and inductor. This is magnetic energy when it's here, and a little later, the current goes down to 0 and the voltage at that point is maximum across here. It's electric energy in the capacitor. So it goes magnetic energy over here, a little bit later, half a cycle later, electric energy here. Magnetic, electric, magnetic, electric. Potential, kinetic, potential, kinetic. Exactly the same kind of thing. And in each step, if you look at these waves, they'd better behave like the networks that you know, because without waves, these networks don't exist. [INAUDIBLE] laws didn't descend as a separate set of rules. They come out of Maxwell's equations, or out of the fundamental equations of any discipline, as we'll see later, that you happen to be working in.

So we have a feeling for that. And the main thing-- you see again, we're just dealing with one dimension, and just getting some feeling as to what the energy is doing and how it relates to a simple lump parameter system.

Now what I'd like to do, the final thing in terms of just fundamentals that we need, is to talk about power flow. In other words, when a wave comes along, there's energy that flows across any boundary. If there weren't, in acoustics, you could never get a microphone to work, because it sure takes some energy to move that diaphragm in and out. So the sound wave is bringing energy with it. That energy gets partially dumped into the microphone. In an anechoic chamber, which you'll see on the last day, the energy that goes into the walls doesn't come out again, to a very good approximation. It gets absorbed and doesn't return. But you have to go to a lot of trouble and a lot of money to get rid of it.

OK. So we want to talk about energy, or power, what we call power flow. Intensity is defined as the average energy flow across an area, perpendicular to the wave. Now normally in textbooks, you're going to see this thing have a cosine theta. If the wave is coming along like this and this is just the x-axis, and there's a plane here, and you asked for, what's the energy crossing this plane per unit time, well, clearly, if the plane is tilted, it's the projection of the plane on this axis, that much energy crosses-- here's the total length of the plane. It's its projection on this axis that counts.

And so you have this cosine theta always tagging along in the books, when you see it. And I've always tagged it along, and I'm going to throw it out now. Because every time I tend to drop it halfway through, and then somebody reminds me and I put it on again. But basically we're going to talk about intersections, power crossing a unit area perpendicular to the wave. And you and I know enough that if the surface is this way and you want it per unit area of the surface, it's the projection of the surface on the normal. That's all. Cosine theta, done once and for all.

Now power flow is P. You may remember some of this from your-- well, let me just think where you've had some of this. Force times distance, you know, is work. The integral of F, dx. So force times velocity, you know from physics. You derived it was power. It's force times velocity, is power. But if you want power per unit area, it's force over area times velocity. Force over area for us-- aha, pressure. So pressure times velocity is power.

Now, it's an in-phase component of velocity? No. We're not in a frequency domain at this point. It's pressure times velocity, whatever it is. P of t times u of t. OK. So the power flow in the wave is P times u. And I'm talking rectangular waves, so I'm just leaving off the vector on u. And intensity is the average power flow. Power flow across a unit area.

Now the average power flow across a unit area is intensity. We'll call it I. So this is the power flow per unit area perpendicular to the surface. We know that now. The average value of this is this. So now we need to get the average value of P times u. Just take a plane wave. And then we will generalize that to a wave moving in a medium in which the air isn't necessarily having z0, such as if you were going down a tube and the tube was filled with something down here on this end of the tube, which didn't give a real z but gave some other z. And what we'll see here when we do this now is going to be the fundamentals that I hope you have seen in electrical engineering. I don't know, but if not, in wave theory, electromagnetic waves, if you've taken that.

OK. Now this is pretty easy, because basically you already have a good discipline in going from time domain to frequency domain, et cetera, and representing the P as a basic exponential building block. P to the j omega t, and u is captial U e to the j omega t. So really what we can do, then, is we can express the intensity, I, as equal to the average value-- time average-- of the real part P times u, of course. I'm assuming that P is expressible as P plus-- these are all plus-going waves-- e to the j omega t. P plus of x and s. And P plus-- P of t is equal to P plus of x and s, e to the j omega t. u of t is equal to that. So I can manipulate the Ps and the u's-- these are for the plus-going wave, if you want to say it-- and take the real part in the end, times e to the j omega t, of course, and get it all out.

So for any kind of a sinusoidal function, you should be able to handle these problems. I don't know if you want me to work an example or not. But I just want you to realize that this is OK. Any questions on that? I'll move on.

SPEAKER 8: [INAUDIBLE]

SPEAKER 1: Oh, two? Yeah, that sounds right. There are two of those things. That's OK now, right?

Thank God. We can save a lot of time. I don't really have to do this energy, except for one minute detail. It's all wrong. Why? Nothing that I've written on that board-- well, this might be OK. Why is that wrong?

SPEAKER 9: [INAUDIBLE]

SPEAKER 1: Out of phase? Yeah, but the sin is much bigger than that. Much bigger than that.

SPEAKER 10: [INAUDIBLE]

SPEAKER 1: It's supposed to be what?

SPEAKER 10: [INAUDIBLE]

SPEAKER 1: Well, this is OK. I could express-- I mean, I'm just doing what I normally do. Here's an expression for the time domain one. I'm expressing it in complex amplitudes. Then at the end, I just tag an e to the j omega t on and take the real part, which I did here.

SPEAKER 11: [INAUDIBLE]

SPEAKER 1: To do with what?

SPEAKER 11: [INAUDIBLE]

SPEAKER 1: No, I'm just considering one dimension here. Yeah?

SPEAKER 12: [INAUDIBLE]

SPEAKER 1: Louder.

SPEAKER 12: [INAUDIBLE]

SPEAKER 1: Did you all get that? Somewhere I think you've seen that. Isn't that easy to fall into a trap headfirst, after you've even seen it on your first assignment? That's exactly-- the real part of a product is not the product of the real parts. So there's nothing wrong, if I wanted to represent P of t times u of t as equal to real part of Pe to the j omega t times the real part of ue to the j omega t, where these are functions of x and omega. That's OK. But not equal to real part of the product, real part of P u e to the j omega t. No, no, no, no.

OK. So now what do we do? We want an answer to this thing, in terms of complex amplitudes. I'm just stating that we want an answer in terms of complex amplitudes, because I know what the answer is, and it's extremely useful. But I can't do that. But there is another way to do it. Anybody have any ideas?

SPEAKER 13: [INAUDIBLE]

SPEAKER 1: Yeah, I can-- time average? I'm sorry?

SPEAKER 13: [INAUDIBLE]

SPEAKER 1: Ah. Well, complex conjugate is the key here. The real parts-- I found out I can't I can't use them in the way that I wanted to do, but I want the answer in terms of complex quantities, so if I represent the pressure, for example, as something like this, that's a vector. Well, if I add to that vector the complex conjugate, I now have a real quantity, and I can express that real quantity in terms of the sum of this and this, without using real part, because the sum of this and this is twice the real part. So the thing I can't do when I get to non-linear operations is use the real part in the way I used it before.

So I'll just express P of x and t as P plus-- I'm going to drop the x and s's as I go along here. I've been threatening to do that for weeks, but I didn't. But I'll write it down the first time. P of x and s, e to the j omega t plus p conjugate-- the conjugate of this complex quantity-- of x and s, e to the minus j omega t. Nothing wrong with that. I haven't used the symbol real part-- oh, I have to take half of it, because the sum of this plus that is twice the real part of the one. And I can do the same for u. u of x and t is equal to 1/2 P plus over z0, or z. I'm going to write it as z now, because I don't care whether it's z0 or not. We'll be talking about z's in general. But what the z is, is P plus over u plus. That's the key. You know that, just like it was for-- if it happens to be in the air, it's z0.

I'm going to drop the x and s's now. e to the j omega t minus p-- sorry, plus. These aren't transmission line equations. P plus over z, conjugate, e to the minus j omega t. Now to multiply P of t and u of t, which is what I want to get power flow, is easy. I can just do this in four parts. Let's see. I will move it up a little bit. So P of x and t, u of x and t, is equal to-- multiply this whole business together, I get 1/4. Now if I take this term times this one, the e to the j omega t's go out. This term times this one, the e to the j omega t goes out. So let's do that first.