Amar G. Bose: 6.312 Lecture 09

AMAR G. BOSE: Today you can just, as far as I'm concerned, take your notes and put a big X through what we did at the last part of the hour. One day, when you get to teach, what you'll find out is that, even though you're not looking at any individual, you sort of scan. And you can see when people are understanding something and when it's not going in. And so I was clear that I wasn't doing a good job.

And I think I'll handle it a little bit differently. Also, because it's at the beginning of the hour, we won't have to rush. I hope we'll get down to it as far as this one, though. So we'll start over on intensity.

Power flow through a unit area perpendicular to the direction of the wave, we'll call it W. I'm going to try something today differently than last time. There is a risk in it, that I can introduce a confusion factor later. I'm going to try it and see how it works. I'm not going to assign any coordinate system here. I'm going to just say space and time.

That's true for the pressure wave. That's true for a velocity wave. It's true for anything. It's a function of space and time.

So W of space and time is equal to pressure times velocity. These are time functions of space and time. Let me write it once, anyway. Unfortunately, I can't abbreviate that by s and t.

Now we want to get expressions, ultimately. This is of interest too. But we want to get expressions for intensity, which is defined as the time average value of this. But let's get this expression first. And then we'll be able to take the time average pretty easily.

And the other thing we know in the beginning is that we would like to get the expression in terms of complex amplitudes. And we found out last time-- don't put the X through that, if you haven't already-- that we couldn't just represent this is as a complex real party, cap P to the j omega t and similarly for u, and then take the real part of the product at the end. It doesn't work because it's a non-linear operation.

But we found out that there's nothing wrong with taking any complex number, representing it as cap P or whatever it is, complex, and then adding its conjugate to it. And you get 2 times the real part. And so that's what we'll do, or what we did, even last time.

We'll represent this P as 1/2, otherwise, the expressions just very long. I'm just going to use cap Pe to the j omega t for this and similarly for this. And I'm not going to say of space and time cap P.

So Pe to the j omega t-- everything I'm writing here in the brackets is for the pressure-- plus the complex amplitude. I'm sorry, plus the complex conjugate of that, e to the minus j omega t. This is the complex conjugate of that. 1/2 of it is the real part. So I have represented this, really, as the real part without using the term real, the expression, times 1/2 Ue to the j omega t plus e to the minus j omega t.

OK, so now we have to just take care of these four terms, which we did also last time, and we get 1/4 out here. I'll take this term times this one because the j omega t's go out that way. PU conjugate plus this term times this one, P conjugate U, times the 1/4, of course, came from the two halves there.

Now we have left this term times this, PU e to the j2 omega t. The exponents add. Plus these two, P conjugate U conjugate e to the minus j2 omega t for all of that.

SPEAKER 1: [INAUDIBLE PHRASE]?

AMAR G. BOSE: Middle?

SPEAKER 2: Middle bracket.

SPEAKER 1: Those brackets, they should all be added?

AMAR G. BOSE: Oh, oh, oh, oh, oh, yeah, sorry. Yeah, thank you. I'm doing the multiplication. OK. And we look and we say, aha. This whole thing here is the conjugate of this. This complex function is the conjugate of this.

And so we can simplify all of this by recognizing that and saying that W of space and time is 1/4. Sorry, if this is the conjugate of this, the sum of them is twice the real part. So I can 1/2 real part of PU conjugate.

Or for that matter, I could say 1/2 of P conjugate U, either one. They're equal because the real part is equal. Plus 1/2 the real part of this one, PU e to the j2 omega t.

Well down to this. We have our expression for power flow, interestingly enough, in terms of complex amplitudes without having inserted the real parts, which I couldn't do because I would wind up by taking the real part of the answer, which I didn't do, didn't have to. I would violate the real part where product is not equal to the product of the real parts.

OK now, to really get a little feel for that, let us take an example. Now what I'm going to do-- I actually did it and it didn't raise any objection already. I just want to see how much of a problem I caused. There's something that's really wrong in this thing.

I didn't cause any problem. Maybe I shouldn't say what it is. This is really a vector, huh? Velocity is a vector.

I've just said the function of space and time. The velocity could be going any way. And the complex amplitude is a vector of U.

Now what happens is as follows for what I want to do in the end. I want to get a relationship between power flowing through an area out here. Let's say the wave is going this way. There's a pressure here, and I want to find out what power goes through a unit area perpendicular to the velocity.

So if I arrived at this spot, I could say, aha, I'll define that as the x-axis. And this is the way the wave is going. And my pressure, of course, is a scalar. And any products that I have to do with velocity and pressure, I can take right there as if this was the x-axis. So I can take a reference axis in space equal to whatever that vector is.

And so, with that attitude, I could have started off with vectors everywhere here, and then got down to here and said, OK, I'm not going to worry about this because the only place I'm going to use it is to compute power at some point. And I can make believe that my coordinate system had this axis along that direction. And whatever wave front I have is all I have to deal with.

OK, so just that you know that that happened. Now to really illustrate this and to make, again, a tie-in between lumped parameter networks and waves, let's-- this P, by the way, P and the velocity, they don't have to be progressive waves. They could be anything.

When I started to give you to do a derivation last time, I thought well, I'll make it easy from the beginning and I'll deal with just plus-going waves. But you don't have to do that. These are whatever waves are there. This is the pressure.

If this was a one-dimensional case, this pressure could be the pressure that was due to the plus-going wave and due to the minus-going wave. Similarly for the velocity. So I would have a relationship, in any case, between pressure and velocity as pressure is Z times velocity by definition, OK? That's what we mean by impedance.

If I let Z be Z magnitude and e to the j some ata, some angle, then P is equal to Z magnitude Ue to the j ata. Now let's use this expression for P up here. And we get W of space and time is equal to-- oh, never mind. I want to use it here.

Putting in for P, Z times U. Well U times U conjugate is U magnitude squared, OK? U, U conjugate is the conjugate of that. When you multiply two complex numbers, you multiply their magnitudes and you add the angles, the angles add out. So this then becomes 1/2 the real part of P. I'm going to use this expression. So it's Z magnitude, U magnitude squared e to j ata by just substituting this in for that.

Similarly, I go down here, this term, 1/2 real part of P is Z magnitude U magnitude squared, which is U times-- well, it doesn't matter. Here, it wasn't even a conjugate. U-- Oh, oh, oh, oh, oh, oh, oh, oh, one more assumption, U equals the reference. I can take that arbitrarily, which means that U equals U magnitude.

The angle of U I'm taking as zero. I can choose a reference. I can nail one quantity, one voltage or current or pressure or velocity, and say OK, that'll be my reference. I get all the rest relative to it. That's all I need. So here, then I would have U magnitude squared again because U is U magnitude-- and U conjugate is U magnitude too-- and e to the j2 omega t plus ata. OK, that's just directly from plugging this expression into here, coming down to here.

Now interesting thing. Well, let's just write that out. 1/2 the real part-- well all of this is clearly real out front-- Z magnitude, U magnitude squared, cosine ata. The only thing that isn't real is this. And the real part of e to j ata is cosine ata. And over here we would have 1/2 Z magnitude U magnitude squared cosine of everything that's up here, 2 omega t plus ata.

OK now, a drawing is in order. This is a constant. And this height, here, of that constant is 1/2 Z magnitude U magnitude squared cosine ata. What we're plotting here is going to be our W of space and time. And this is time.

Now we have, aha, a cosine function of the same amplitude as this, as the same amplitude as this, OK? Now cosine theta is equal to or less than 1. So let's say, if it's somewhat less than 1, I'll draw it this way and so on. And this amplitude here, from here to here, is 1/2 Z magnitude U magnitude squared.

So this one is equal to or less than that, which means that, if we take a look at different cases-- for example, let's suppose that ata, the angle of the impedance, was zero. It's resistive like it is for a plane wave, let's say-- then this curve would look like this because the average, the DC, would be just equal to the peak value of the cosine.

So that's for ata equals zero. If ata equaled 1, we'd have something like this. Pardon me?

SPEAKER 3: [INAUDIBLE].

AMAR G. BOSE: That's all right, pi/2, ata equals pi/2. Thank you. Namely, ata, the angle equal to pi/2, would be a totally reactive load. That would be an inductor like an inductor or a capacitor.

And then what W of t would look like-- we'll tie this down, this is a little bit abstract in terms of waves yet, but we'll get there-- would look like this. And for something in between, it will look like this.

Now the interesting thing is that it goes below zero in all cases except for ata equals zero. And going below zero means that power is being returned from that point in space. But we can look at that point in space, when the wave is going like this, it's just like it's seeing two terminals on a box.

And power comes out of the box and goes in. That's exactly what you would expect if you had a capacitor out there, a reactive load or an inductor in there, no resistors. Or in fact, a whole bag full of capacitors and inductors all interconnected still is a reactive load. j, the impedance is always imaginary.

So power, in that situation, will always look like this. And just as much power returns as goes in. And it'd better because there's no means of dissipation. If this thing is full of all sorts of reactive elements, there cannot be any average power.

By the way, there's a term that you may-- I don't know. How many of you have heard of vector power? Oh, boy, one or two. Huh. There is a term which this is suggestive and that's how it came about, that look, the real part of this vector, PU conjugate, is average power.

So let's talk about vector power. And we'll call vector power 1/2, if you want, P times U conjugate or, in electrical engineering, V times I conjugate. Or it doesn't matter which way you want to do it, the sign will only change. You can make the conjugates interchange.

You can take this as your vector power or this as your vector power. The real part is the same. That's the only thing that has real physical significance. But the angle of this quantity is, as we have seen, the angle of the impedance ata.

And so some people, especially in power engineering, think conveniently in terms of vector power. And they know that, if that vector is real, in real axis, then the load is resistive. If it's anywhere else, the load is reactive. One way it's inductive. In the other direction, its capacitive.

And so that's just a term that, in electrical engineering, you see a lot of times. But its motivation came strictly from-- I shouldn't say these equations. But as we know, the transmission line equations, you can get out in voltage and current exactly the same as this. And if you like, this becomes voltage in this current. Or if you like, this becomes current and then this voltage.

OK. So now, we mentioned at the end of the last lecture that the power company doesn't like you if you have reactive loads. Well look, for the moment, you can think of this as current, if you want. Just think of this as current. For the electrical engineers, it'll be easier.

The power company is shipping down its lines to you a peak value of AC current equal to this much. Depending on what you do in your house, you can take this much power out. That's how much you can get out of it, if cosine ata was 1.

By the way, this thing is called, in electrical engineering, power factor. It becomes 1 when ata is pi/2, when the-- sorry, when ata is zero, this cosine of zero is 1. And when ata is plus or minus pi/2, it becomes zero. So for low power factors, you are taking very little power out.

Now you might say, so what? Well your meter in the house reads the power you're actually taking out of this, the average power it reads. So now why do you think the power company is upset with you if you have a low power factor? Ah, OK, go ahead.

SPEAKER 4: [INAUDIBLE PHRASE] down the wire, you should have resistance there [INAUDIBLE PHRASE].

AMAR G. BOSE: That's right on, yeah. He's a practicing engineer. That's why I was hesitant to-- he's a graduate engineer and so I was hesitant to call on him. But it's exactly right. Look.

Between the power company and you is a big transmission line. And that's a giant resistor. And so the power company is heating the world out there when he sends power to you.

And why does he care that you only take a little bit out? Because you only pay for a little bit, and he has to still heat the rest of the world. So he's delivering this much current, which is I squared R loss in the lines, and you're paying for this little, tiny bit of power out here. He doesn't like that. Understandable.

If you were taking this much power out from it at the other end, he's happy because his rates are based on you doing that. Normal home, electric light bulbs are dominant. So he sends you a big bill each month if you're doing that. And he sends you a very little bill if you're doing that. And, by God, he has the same losses in his lines, so he's not very happy.

So the power company decided a long time ago that, if you had really reactive loads, they were going to visit you in your shop and look at what you were doing. And they would say, well you have a very low power factor. So I'm sorry, per watt, we're going to charge you a lot more for your bill, yeah, on your bill.

SPEAKER 5: So for facilities that have this large [INAUDIBLE] factor, can they just locally install some device that would be able to [INAUDIBLE]?

AMAR G. BOSE: Aha! Can they install a device in which the power company wouldn't know the difference at the other end? Well suppose you had this problem. Here was your shop. And let's say you have some load that you're taking out. With all these machines, you're doing some work, but you have this inductor across here, which is the inductance of all the loaders and whatnot.

And you know that you're going to get socked for this. You're going to have a new rate for the per watt that you use. Is there anything you can think of that you might do? That was the question.

SPEAKER 6: Put in a capacitor?

AMAR G. BOSE: Aha! And how would you pick the capacitor? I can take a 0.01 capacitor, it'd cost me $0.05 or $0.10, I can stick it across a line. I know the thing isn't going to do much good.

SPEAKER 7: Make the resonance [INAUDIBLE]?

AMAR G. BOSE: Sure, you make the resonance at 60 Hertz. What's the impedance of a parallel combination at resonance?

SPEAKER 7: [INAUDIBLE].

AMAR G. BOSE: Wait a minute. What's the impedance of a parallel combination at resonance? Any parallel combination at resonance? Oh, we're not going to finish. Somebody says infinity.

Wait a minute, let's do something. Let's talk about the admittance of this thing. The admittance of this is Y equals 1/Z equals I/V.

And we know that the VI relation of this thing is V equals 1/J omega C times I, OK? So I/V is equal to J omega C. Plot that in the S plane, and it's 1/J omega C is a vector pointing up. Sorry, J omega C is the vector pointing up.

Now the inductor of VI is V equals J omega L. So the admittance y Y is equal to 1/J-- V is J omega L times I. The admittance, which is I/V, is 1/J omega L, which is a vector pointing down.

Now the sum of the admittances of two things in parallel, whatever they are, is the admittance looking in here, OK? Because you're adding the currents, so you sum admittances to get the admittance looking in here. And if you want the impedance, upside down of that.

Now look what happens. You start off at very low frequency. Well the way I've drawn them, let's say I start off with high frequency. This is much larger than this.

So the resultant at any one of those high frequencies is this minus this, which is this. It looks like a smaller capacitor at this particular frequency. In other words, if I choose a frequency, the one for which these are represented, the resultant of these two vectors is a vector that comes there.

If I decrease the frequency further, this factor comes down and this one gets larger. And eventually, they're equal. When they're equal, guess what? J omega C is equal to 1/J omega L, defined as omega is equal to 1 over root of LC. That's your resonance frequency. Then the resultant vector is zero, the two of them. Then it's a infinite at that frequency.

If the sum of the admittance is a zero, the impedance is infinite, OK? Below that frequency, this vector becomes smaller and this one gets bigger. The sum is then this minus this, which is maybe here. What is that? That's an inductor of a larger value than the real inductor.

So at any one frequency above resonance, this is the capacitor. Its value changes, depending on the frequency, when I talk about it, at any one frequency below resonance and inductor. So at the 60 Hertz, you put this thing in here. And now the power company doesn't know the difference. Now would you really think that's cheating?

There are a lot of companies that have this. If you go in, right where the power lines come in, you'll find this huge bank. It turns out that 60 Hertz, to resonate the inductances of machines, it's not that little 0.05 capacitor that you have in your pocket. It's banks of them sitting all over the floor, and they have it.

Now is that cheating? How many think it is? Nobody. It's not cheating.

Why is that cheating? Because it's exactly the same as your house now. And the power company-- now you've changed this power factor to unity. So you are now, for the same amount of current that he's delivering, you're taking as much as you can get out of it. And your watt meter is reading that much. You're paying for what you're using.

In other words, all your rates are based upon having a resistive load because, when you have a resistive load, the electric company's already thought about that, what his average line length is and what not and how much power he wastes in heating the lines. So as long as you take that much out and he can bill it to you, he's very happy. So you can even buy the capacitors from him.

So when you think now of reactive impedances, what you always should think about is power going in and coming back. Whenever there's any kind of reactive component to the impedance, there's a sloshing back and forth of this. And we will see, even for example, because I've done this generally this time, this Z doesn't have to be real, even in air.

The only place Z was real in air was a plane wave. As you've already sort of seen, I think in homework, but we'll see in much more detail, you have a spherical wave, and unless you go very far away from where it's being launched, the wave is not a plane wave. And the impedance is not real. It has a reactive part.

So believe it or not-- and we'll see this when we come to it-- you have a pulsating sphere here. When you measure over here, then you'll find out that, through this in this field, energy is sloshing back and forth. The impedance is not real. And so you have this power going back like that.

OK, wow, big detour. Any questions? OK, I forgot what we're going to do now. Oh yes, generalized elements. Why generalize an element? Simply because, again, the incredible overlap, if you wish, of the different disciplines governed by the same simple mathematics.

You have elements in electrical engineering that you've heard of. You have elements in mechanical engineering that most of you heard of. You have elements in acoustics that you will hear of.

Whenever you have systems governed by constant, coefficient linear differential equations, you have a network theory, if you want, for that discipline. And you can swap between disciplines, as we will begin to see. This is marking the transition in the subject in which we'll be now, mostly, building insight and not too much new analysis or mathematical-- well analysis, yes, but mathematics, no.

Let's see, for the moment, when we talk about generalized elements, just think of it mathematically. Don't try to associate this in your mind, for the first few minutes, as any element that you know. We will assign a pair of numbers associated with this symbol.

And only from bias, we will call this a. And we'll put a plus and a minus sign. We'll draw an arrow here. We'll call this b. And we'll say that there's some property of the element, which we will call c. And c has something to do with the relationship between a and b. It's a generalized element, OK?

Now there are, it turns out, in any of the disciplines, if you consider three elements, three different kinds of elements and a source and another thing-- called, again, with bias-- a transformer, then that's sufficient to make up the whole body of that discipline, just these four things, three elements, sources and transformers. There's another thing you can do. There's another alternative to that also, but let's just focus on this one.

A source-- by the way, well we have two sources, two symbols for them. One is this. And I'll explain. Whoop, let me just use a different number here, a different value.

Now the three kinds of generalized elements that we have in these different disciplines that are governed by the constant coefficient lum parameters are the following. a, copying that down, a equals b times c or c equals a/b. By the way, I don't think I-- did I mentioned this about the--

You can imagine this now, two MIT graduates. This was in the first days of the company. The company was now big enough to hire a third person who was going to be a technician to one of the engineers. So these two MIT fellows interviewed the-- this is true-- they interviewed this technician.

And then they came back. And their eyes were about this big. And they told me that evening. They said, gees, we interviewed this fellow and we got an amazing result. And he was telling us how much he knew. And that, for a resistor, he knew all three Ohm's Laws, V equals IR, I equals V/R and R equals V/I. And he was very happy about that.

Now that is a very interesting lesson, that the more mathematics you know, the less laws you have to remember, all right?

[LAUGHTER]

AMAR G. BOSE: I'm going to write these things as the inverse because we're going to use them that way, not because you couldn't calculate them, all right?

[LAUGHTER]

AMAR G. BOSE: So a, b, c. Another thing we have is a equals c db/dt for a generalized element. Or c equals 1/-- sorry, the variables associated with the elements are the across-- I should have said that to you-- across. This is the across variable. And this is, so-called, through variable.

And so this is then the inverse of that b equals 1/c integral i d tau. And the third one being a equals 1/c integral bd tau. And b equals c da/dt.

Now what distinguishes an element from a source? It's very fundamental. An element describes a relationship between the two variables, a and b. There they are.

A source constrains one variable. It does not relate the two. In other words, this symbol means that the across variable, in this case the little a is equal to big A independent of what the through variable is.

You'll have a through variable there. It doesn't make any difference what it is, this is A. So it does not define a relationship between the two variables, an element does. This symbol means that the through variable is equal to cap A no matter what the across variable is. That's the definition of a source in the generalized elements.

Now we just have transformers left to put down. A generalized transformer, again, you can see the bias of electrical engineers in this, but that's life. b, let's call it, g, f and c, c has to do with that thing that gives a relationship between the variables just like it did up there. Now this is just a symbol, it doesn't mean anything. Don't try to related it to what you may be thinking.

And this symbol means the following thing mathematically, namely that a is equal to c times f and b is equal to 1/c times g. With the signs as I have, when you put arrows this way, this means if b is positive, the through variable, whatever it is, is that direction and, similarly, there. Yeah?

SPEAKER 8: On the board above, for the second generalized element, [INAUDIBLE] the standard you're using, you want to say b [INAUDIBLE] a and b?

AMAR G. BOSE: This fellow?

SPEAKER 8: Yeah. [INAUDIBLE].

AMAR G. BOSE: Oh, oh, what did I do? Oh, boy. The bias really came through here.

[LAUGHTER]

AMAR G. BOSE: Woo, OK. Yeah, 1/c integral i dt is something that we all know too closely. Let's see, OK, now just take a look at this.

This is the definition. But look how much you can get out of this little definition. Suppose I multiply a times b. If I do that, I multiply the right-hand side of the equation. The c's go out. I get f times g.

Now in all these different disciplines, I've already told you that one of the variables is going to be whatever it is. Sorry, the product of the two variables is going to be power. And so what does this say? Instantaneous power in is the instantaneous power going out. So it says two things, just from that simple thing.

It says there's no storage in there. These are instantaneous quantities. Whatever that symbol is there that we put in the box, it can't store any energy. It can't dissipate any energy, two things, because the power, at any time going in, is equal to the power coming out.

OK, now we can go to first electrical engineering. The first one up there, I think we all know that c is a resistor, R. The through variable that we would normally use is current. The across variable is voltage.

Network theory, by the way-- and you'll see this in a matter of a week or so-- it doesn't have to be this way. It's just a natural that the across variable you call voltage because you measured it across. The through variable you call current because you measured it through. You can make circuits in which these things are all reversed, it doesn't matter, not the physical things, but the circuit.

So V equals i times R. And the other Ohm's Law, i equals V/R. Now we have, let's see, the next one. Cross variable is something times the derivative of the through variable. What element would you like to make there? Huh? I'm not going to go any further than that. Across variable is some constant times the derivative of the through variable. Inductor? Capacitor?

SPEAKER 9: [INAUDIBLE PHRASE].

AMAR G. BOSE: OK. We made it. All right. vi V equals L di/dt. i equals 1/L integral V d tau where c now became the inductance L.

And finally, the last one, of course, has to be a capacitor, vi. v is the across, i is the-- that is another C. It's a big C. So we would have here, V, the across variable, was 1 over the capacitance integral, i d tau. And i equals c dv/dt.

OK, so much for the electrical elements. Let's go to the mechanical ones. First element, a equals bc. I will ask, for the moment, the mechanical engineers to just smile and keep quiet. I want the electrical engineers to tell me what the first element is in mechanical engineering.

SPEAKER 10: Dashpot.

AMAR G. BOSE: Dashpot. What's a dashpot? Do you know?

SPEAKER 10: [INAUDIBLE].

AMAR G. BOSE: It's right, but what is it? You have damping, that's right. It's got to be because a times b is c. And c is going to be some real number.

SPEAKER 10: Shock absorber?

AMAR G. BOSE: Shock absorber, absolutely. What is proportional to what in a shock? I don't know if any of you've gone into a store. Some stores have promotional things to sell shock absorbers.

And they have a bad one hanging on the wall, and then they have a good one. And if you try to push the bad one, it just goes like that. And you try to push the good one, and it's a different story. So what variables are involved in the mechanical elements that the normal analysis of mechanical--

SPEAKER 11: Force.

AMAR G. BOSE: Force--

SPEAKER 11: Velocity.

AMAR G. BOSE: --and velocity. Those are the two. It doesn't have to be, but it is a natural just like in electrical engineering. You usually think-- we're going to see circuits that do all sorts of interesting things-- usually think of velocity as the across variable.

You grab this shock absorber-- it's a cylinder about so big and two terminals on it-- and you push it. The velocity is the across variable. And the force is the through variable. And the shock absorber, with viscous damping force, is proportional to velocity.

So this element-- mechanical engineers have a funny symbol for this thing-- is that. Now this is force. And this is velocity-- oops, a u-- velocity. This symbol isn't so funny, really, because, if you imagine that thing filled with oil and a top on it somehow, that would do it.

And that's actually what a shock absorber is. It forces oil through a lot of little orifices inside there. And if the shock absorber leaks and the oil goes out, that's when you have this thing where you see the cars going along the road like this.

[LAUGHTER]

AMAR G. BOSE: So this u is equal to r mechanical. And usually the people denote this, at least in acoustics, when we get to it, the velocity related to force in this fashion, velocity is something times force. It's a little r.

When they do the other one, the inverse of that, of course, we can write it this way. But believe it or not, how it's very often written is-- just so that you know this, I don't care what we call it-- but they call it a big R. Force proportional to velocity is the viscous damping. But big R is equal to that. It's the same as in electrical engineering. We sometimes call 1/r the conductance, g. Yes?

SPEAKER 12: The force is the across frame?

AMAR G. BOSE: No, in that it'll turn out-- and if I get to it, it'll turn out soon-- that the natural, not required, but the natural thinking, if you have a mechanical kludge here, you look at the elements, you see velocity, one term. If you take a spring, for example, you think you're measuring velocity of this terminal relative to that terminal. And the force is something you would measure in the element, like with a strain gauge. So the normal, you think of it velocity is the across frame.

OK, now the next one, anybody know what that is? Well give me another element in mechanical-- I won't ask you to look at those equations and tell me what-- give me another element in mechanical engineering.

SPEAKER 13: Spring.

AMAR G. BOSE: Spring, all right. They copied, in this case, the symbol, f u. Now this fellow, how do you relate f and u in a spring? Mechanical engineers, still quiet.

SPEAKER 14: The f?

AMAR G. BOSE: F, yeah, but that's not relating f and u. f equals kx. Yeah, you go over here on the side, f equals kx. And k is sometimes called 1 over the compliance-- sub m means mechanical to us-- and 1 over the compliance times x. But if you wanted to get that in terms of u, 1 over the compliance, how do I get that into u?

SPEAKER 15: Integral.

AMAR G. BOSE: Integral ud tau. So f is equal to 1/cm integral ud tau if I'm listing. So u would be equal to c sub m mechanical compliance times df/dt.

And this is the other one, of course, f equals 1/c sub m integral ud tau. Don't try to remember any of these things. It's hard, sort of hard, to forget this one. But if you happen to remember that, you're in.

And finally, the last one up here is a mass. Now, if this were a rigid body, this is a mass. There's just one minor problem. Where's the other terminal?

[LAUGHTER]

AMAR G. BOSE: Oops.

[LAUGHTER]

AMAR G. BOSE: Where's the second terminal? I've got it on the shock absorber. I've got it on the spring. Where the heck is the second terminal?

So I guess we can't use this thing. It doesn't fit into our network theory. You've got to have an across variable. You've got to have a through variable for everything.

There is a second terminal. Maybe you're too far away to see it, but there is one. There's a second terminal in that picture over there too. Any ideas at all? What?

SPEAKER 16: [INAUDIBLE].

AMAR G. BOSE: Yeah, [UNINTELLIGIBLE]. If I were standing on ice here, if I went like that, I would go that way. The fact that there's friction down here, I'm pushing, when I move this thing relative to down there, I am pushing this against the inertial system, whatever inertial system I'm considering.

The second terminal on this mass is earth, is ground. And that's why the mechanical engineers have this for a symbol for this thing. I guess it's like this. And I think they use a ground like that.

And you can still have the force as the through variable and the velocity as the across variable. But one side of this is always going to ground. And now you understand how disadvantaged mechanical engineers are.

Electrical engineers can put capacitors anywhere they want. They can put them between nodes. They can put them from nodes to ground. And the poor mechanical fellow's got to put every mass in their network, that they design a complete filter, the mass has got to go through earth to ground.

It's like saying, oh boy, you have a network theory. But every time you want to use a capacitor, make sure that it goes down to the ground. It doesn't give you too much freedom, does it? It's amazing they do as well as they do, you know? That's a big constraint.

Now the relations for that, where do they come from? Newton's Law, force is equal to mass times acceleration. Well acceleration, I wanted velocity, so du/dt. And the inverse of that, u is 1/m integral fd tau.

So now we have those elements. However, that is not in the-- I just told you-- oh, oh, oh, I didn't tell you anything about transformers. Transformers for each discipline, that's pretty simple at this point. And we'll get to it.

But in electrical engineering, at this point, you recognize this as a transformer in which the voltage at this terminal here, V1, is the transformer, usually, c becomes, at t it turns ratio times V2, if this was c to 1. And i1 is 1/t times i2, i2 being the through variable over here. You know what that is.

In the mechanical side, if you had a transformer like that, you would say, for your cross variable here, U1 is equal to whatever that c was, c times u2, the across variable on this side. And f1 is equal to 1/c times F2, transformer.

By the way, you know what the transformer is in electrical. What's the transformer in mechanical? A lever is a good idea, a gear.

Just take a lever-- maybe you can do that in recitation-- take a lever, assign a force and a velocity to each end and write the equilibrium equations and you get the transformer relations. And it's clear, if this is a massless lever, that there's no energy stored in it or dissipated if the pivot is good.

So we have that. But all of this isn't enough for a network theory, for any kind of a network theory. There's something else we need. Anybody have any idea what you need to complete a network theory on any of these disciplines? I won't do the acoustic elements now because we haven't derived them all yet.

SPEAKER 17: Sources?

AMAR G. BOSE: Oh, sources? Yeah, sources. OK. This, you know is a voltage source. If you put this symbol in and constrain the across variable in electrical engineering, this is a current source. In mechanical engineering, this is a constant velocity across here by definition. This is a constant force in here equal to whatever the symbol is that's out in front.

So we have sources. We have transformers. We have the three elements. And it's not enough in any discipline. Yeah?

SPEAKER 18: Something like wires?

AMAR G. BOSE: Yeah, what do wires do?

SPEAKER 18: They [INAUDIBLE PHRASE].

AMAR G. BOSE: They connect things together. And now, when you connect things together, there's another set of constraints that enter. Electrical engineers, what are they?

SPEAKER 19: Kirchhoff?

AMAR G. BOSE: Kirchhoff's? OK, let's see. We have the elements. Let's see, Kirchhoff laws, if I remember correctly, they something like this.

If you had a current coming in here, i1, maybe a current positive-- arbitrary directions here-- i2, i3. And Kirchhoff's current law, KCL, would say summation of the currents coming into a node or summations of currents going out of a node-- that's the node-- is equal to zero. Summation of ik, whatever you want to call them, is equal to zero.

Kirchhoff's voltage law, I think you all know would say that, if you had voltage drops around this V1, V2 around a closed path-- these are wires and elements. An element might be here. Another element, it just might be an r and a c and an l, are connected all together at these nodes, V2, V3 plus, minus, let's say, minus, plus.

And this would say that-- which, by the way, in this example, this thing, of course, reads-- let's suppose I take currents going out, just arbitrarily, that would be minus i1 plus i2, plus i3 is equal to zero. That's the statement for that. And this, you draw a little circle, and that means that you're taking voltages around a closed path. And this would be sigma vk equals zero is KVL.

And in this particular case, depending on-- I can take voltage rises around a circle. I can take voltage drops instead, long as I'm consistent. I can go any direction I want. I drew the thing counterclockwise, so let's go counterclockwise.

This is a voltage rise. And I'll take some of the voltage rises as zero. V3, if I started here and went this way, I'd have V3. That next one is a drop minus V2. The next one is a rise plus V1 is equal to zero, KVL. Do you know where these came from, by the way?

SPEAKER 20: [INAUDIBLE].

AMAR G. BOSE: Yeah, have you seen that? I really don't know how you were introduced to Kirchhoff's laws, as laws, or as something that isn't true when it comes from Maxwell's equations?

How many of you have seen the Maxwell equation derivation of almost this? Oh, gees, only one or two. OK, I will spend a couple-- oh, boy-- I will spend a couple of minutes and do it because there's a very interesting thing to be learned from this in terms of modeling.

Let's see, what do we have here? Oh, divergence of D, the displacement vector, is rho. Divergence of B is zero. Curl of H is conduction current plus displacement current derivative, which is derivative of the displacement vector. Curl of E is equal to minus partial of B with respect to t.

Let's take a look. Where did we start up here, KCL? KCL says the sum of the currents flowing out of this node here are zero. Let's think of the node like so. Divergence of the curl of a vector-- that's one of the vector relationships-- the divergence of the curl of any vector is identically zero. So divergence of the-- By the way, I'm not going to test you or anything on this, but I'm just doing it as a little side. If you see it, it helps build something for modeling for you.

Divergence of the curl of H is equal to zero. And that is equal to divergence of J, the conduction current, J sub c, plus the divergence of this fellow. But divergence is a spatial operator, so I can take it inside the partial, so a partial of respective t of divergence of D.

Now divergence of D I can take from here. So I get divergence of J sub c-- this is all equal to zero, so I'll take this on the left-hand side and leave this on the right-- is minus partial of rho with respect to t.

So it says divergence is partial of rho. That's not right exactly where we would like it. We have a little volume here, the node, where all these connectors are coming into it. So I'd like to get this thing into-- well maybe I should do it and then explain. Hold it.

Divergence of J sub c integrated over the volume of this node, over the volume of that node, is equal to the vector integrated J sub c over the area. This we've seen already long before in the subject, the J sub c in acoustics. J sub c times dA over the area of this node which the connecting wires are coming into over the area, and that is equal to--

Well I've taken this and integrated it over volume. And this charge I will now integrate over the volume to get the total charge. So that is D-- how shall I put it-- D RHO, big RHO or something, Dt.

Now this is a big RHO. I don't like to call it RHO zero because we've already used that. But I've just integrated both sides over the volume and then reduced this to an area integral.

Now current density integrated over an area is the current. So this comes down to sigma i sub k in the case of the current i1, i2, i3, et cetera, going out there. This is sigma i sub k, but the darn thing isn't equal to zero. It's equal to the rate of change of rho of the charge density on there. So the assumption to get to Kirchhoff laws is that this fellow is very small.