Amar G. Bose: 6.312 Lecture 11

AMAR BOSE: Today, I thought I'd talk a little bit more about these loop and node equations, and just give you an indication of how powerful it is in the sense that you made be analyzing some physical device. It could be a the first time a transistor was developed, and you get a set of equations, and then you make a model from those equations. Or it could be a control system. It could be anything. So what I'm going to do is I'm just going to make up some equations on the board, and then ask you some questions about what the network is. This will take a little time. I have to be a little bit careful, and it's not obvious to you why I have to be careful at the moment, but it will be later.

Let's see. OK. Let me just see. I'm going to-- if I haven't been careful enough, it'll be obvious in a minute. First question, you can already start thinking about. The first question I'm going to ask you is, just by looking at this set of equations, what can you tell me about the network, before you start drawing anything, before you start doing anything beyond just looking at the nature of the equations?

[TALKING TO SELF AT CHALKBOARD]

OK. These may or may not do what I want yet, but we'll see. And we'll adjust them if it doesn't. What can you tell me about the nature of a network that would be yielding these equations? First of all, what kind of equations are they? Loop or node equations?

MALE SPEAKER: Node.

AMAR BOSE: Node, because you have the voltages at-- how many nodes are there?

MALE SPEAKER: Three.

AMAR BOSE: Three. OK. What kind of elements are in the network? I'm going to make it bigger. And of course there's a datum node here. What kind of elements?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Inductors.

MALE SPEAKER: Capacitors.

AMAR BOSE: Capacitors.

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: I hear capacitors very frequently. How many say there's capacitors in it? How many say there's inductors? How many say there are resistors or conductances. Let's see. If there were capacitors and inductors, whatever these equations are, whether they're loop equations or node equations, you would find some terms that look like j omega times something. And you'd find some terms that looked like this, because if they're both, this has got to happen. Now I don't see any 1 over j omega kinds of terms. So there's only two kinds of elements in this thing, conductances and capacitances. Because the impedance of a capacitance, 1 over j omega c. An admittance has to multiply a voltage to get a current, so this is the admittance of something.

OK. Now let's see if we can make the network. Anybody want to connect something to node one? What's the easiest thing to connect to node one, looking at this equation? Current source. OK, that goes node one-- this is the value of the current source. The fact that it's a current source is the arrow. OK, we took care of that one. What else do you want to connect to it?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: 5 to ground?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Now admittances-- OK. Let's go back and review what we saw the first time. The diagonal terms are the sum of the admittances that connect to that node. The off-diagonal terms, remember, are the ones that represent the sum of the admittances that go between that node and the node for which you're writing the equation. In other words, this is written for node one.

Now all the ones that are in here, which is the sum of everything that connects to node one-- if they don't appear somewhere else, then those elements could not go to any other node. They go to ground, we saw last time. So with that much knowledge, we should be able to construct the network. What goes here, and where?

MALE SPEAKER: [INAUDIBLE] 3 to ground?

AMAR BOSE: Admittance of how much?

MALE SPEAKER: 5

AMAR BOSE: 5. You mean a conductance.

MALE SPEAKER: Yeah.

AMAR BOSE: Of 5. Let me just see. Since the resistor symbol is used for the conductance and the resistance, I always have to put there a G equals 5, then, to let you know that that's in mhos, or you've got to put down five mhos or whatnot. Let's see. You got that because you saw a 10 here? How did you get it?

MALE SPEAKER: [INAUDIBLE] so I knew that there was a conductance of 5 when I get to node 2 but then there's a total conductance of 10. Yeah I mistook the 2.

AMAR BOSE: There was nothing here going over into conductance. There's a total of 10 connected to the node. Five of them are going to node two, which tells us right away that there must be a 5 here. So, OK, that's right. Now what else connects to node one, if anything? Anything? I'll tell you what you do. When you're doing this, it's sometimes easier to fill in the mutual terms first. So this is the sum of the admittances that goes between node two and the node we're writing the equation for, node one. We already have this fellow, so what's this?

MALE SPEAKER: Capacitor?

AMAR BOSE: A capacitor of value 4. OK, that takes care of everything that goes between node two and node one. Where does this fellow go?

MALE SPEAKER: To node three.

AMAR BOSE: To node three from?

MALE SPEAKER: One.

AMAR BOSE: One. Yeah. It borders between the node for which this is the variable and the equation for which you're writing that node. So that is a capacitor of value two. Now let's see.

Now let's just check that all of this is all right now. The sum of all the capacitances connecting to node one is 4 plus 2, which is 6. That's right. The sum of the conductances which connect to node 1 is 5 plus 5, is 10. That's right. And so these we already constructed to be correct. OK. So now we have to fill in the rest.

Between node two and node three, what goes? Anybody? Yeah.

MALE SPEAKER: Something with a conductance 3.

AMAR BOSE: 3. Here it is. This is node three. The equation is for node two. So that has a conductance of 3. That's all that goes between those nodes.

Now between node two and node one, you notice I very carefully made this equal to this, as it has to be. Symmetrical, remember? The matrix is symmetrical. You have your diagonal terms, and then your off-diagonal will have symmetry, of course, as it has to be because the admittance that connects from here to here is the admittance that connects between these two nodes, and had better show up on both sides. So that was already taken care of. So this fellow, we know, is right.

Now we already fixed this one up, and so we only have this to look at. This is the total admittance going to that node. So there's a capacitance of value 7 connected to that. And there's only 4 in the mutual, so there must be 5 here, or 4 from 7 leaves 3, I think. So that takes care of that. You had the 4 here, and you had more left over that didn't connect to any off-diagonal terms, so it had to go to ground. OK.

So going down to the third equation, this is what connects node one and node three. We already took care of that. And that comes from the symmetry that you see here. Then we have this fellow here, which we already took care of, connecting node two and node three. And that's by the symmetry. He's taken care of. So all we have to do is look at this. The total capacitance connected to node three must be equal to 2. And in fact, there's no capacitance going from node three to anywhere else, so--

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: What's that?

MALE SPEAKER: Node one [INAUDIBLE]

AMAR BOSE: Node one. Oh, yes, there is. Sorry. 2. So the total was 3, so this must be 1. No. The total is 2, I'm sorry. This is 0. OK. And let's see, the total conductance is 4 going in there. And what do we have so far? Do we have 3 in the other paths? No. So that must be a 1.

So that is the network. You come up with equations for something. It could be magnetics. Well, it won't be magnetics with the capacitors, but it could be any kind of a system. And then you can represent that set of equations here. Yes?

MALE SPEAKER: [INAUDIBLE] node two.

AMAR BOSE: Node two.

MALE SPEAKER: [INAUDIBLE] three and five connected to it, but if you take a look at your middle equation, [INAUDIBLE] 7 plus 3.

AMAR BOSE: Let's just check. 7 capacitance and 3 conductance.

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Capacitance. Oh, 7 capacitance and 3. So 7-- this is a 2, huh?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Wait a minute. Wait a minute. I'm looking at this node.

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: OK, the 3. There's a total conductance of 3, huh? OK, I screwed up. Basically, that's the problem. But that's very good, because I can now illustrate what the problem is. You get a negative conductance going down here. Now it is impossible if you try to do it with a passive network in this structure. Just a couple of minutes spent about this. Then we'll patch it up.

It's possible, when you define a structure like we did with nodes, to come out with negative elements. If you pick the-- well, you don't normally pick these things right or wrong. I just made up an imaginary thing over there, and it happened to illustrate this. But if you have a certain structure, you can come out with negative elements in it. It doesn't mean that there's not another topology for the network that will have all positive elements. There may or may not be such a thing. If there isn't such a thing, then you have to use active networks to synthesize it. The fact that there was a negative element is not a disaster by any means, when you come up with this in your set of equations, because you can realize that with negative elements.

Normally if you're analyzing a physical system that's stable, you don't care because so what? You put the negative element in there. But if you've come up with one that's unstable, you put the negative element in and the thing will take off. But basically, the sum of the rule that we were going to get to, but we got there suddenly, is that the admittances connecting to a node have to be equal to or greater than the sum of all the ones that connect from that node to the mutuals. Otherwise you need a negative element from the node.

Let's just see what happens here, if we can look at this again. This was the culprit. There was a 5 connecting node two to node one somewhere. And there was a 3 connecting it to that. So if the total was 3, and I have 8, I must have a minus 5 down here to make this network realize those equations. So let's for now say we want to have all positive elements. Then I have to boost this fellow up. This is 5 here, 3 here. That's 8. If I put in just 8 here, then I have nothing going down there in conductance. If I put in 9, then I have a 1 going down here. Let's see, was the capacitance OK? There was 7 connected to it. 4 went somewhere else, to node one. And so that should have left a 3 here.

OK. Now let's suppose for some reason that I look at this network, I analyze it. I say, gee, I'm going to make a change in this network. I'm going to put into this network an inductance right down here. This, by the way, is perhaps one of the most complicated things you bump into, a three-node network. They don't get much bigger than this for a certain reason. The reason is tolerances of elements. When you get the huge network like this, it's very hard to manufacture that thing, because just the tolerances can throw off the frequencies enough. So you tend to make little ones, and then you cascade them. So this is a pretty sophisticated network.

So suppose I put in inductance here of value 2. What do I do to the equations over here? Anybody know? I'm now going the reverse way. I'm writing equations to match a network. I had the equations that matched the network without the inductor in here. How do I match the equations now to the network with the inductor?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: 1 over j omega 2 right here. Exactly. Minus 1 over j omega 2, plus j omega 7.

Now a very interesting thing. If, for example, I made this a 10, then this would disappear, right? If I wanted the same equations. So if I put this as a 10, there's 10 connecting between node one and node two. The total connecting to node one, total conductance, is already 10. So this would disappear.

Now what I want to just tell you-- I won't do it for you, because this is another half an hour or so, but there is a very nice way and it's virtually unknown in the industry. It was due to Professor Guillemin here, who was a world authority on networks. And I don't even find it in his books today, but it was just known to the students that he showed it to one day in a graduate class.

There's a way of taking a network and actually changing the values around, sometimes changing the topology without changing the desired transfer function. In other words, you might have as a desired transfer function for this thing the voltage here in response to the current here. And that has a certain frequency response that you want, therefore a certain number of poles and zeros, and these are the equations that govern that. But you, for some reason-- I'll give you a couple of reasons in a second-- might want to change something in this network, some of the values or the topology. And you can, in many cases, actually do that and preserve the transfer function.

Now I'll give you an example. It might be that you'd like to get rid of an element. And I showed you that if this one-- we don't know whether we can do this yet, but if you could change this element and you could get rid of this one-- but that might change the transfer function. It may or may not. You don't know that yet.

I'll give you another example. The first loudspeaker that our company developed was a 901. And it had a thing called an active equalizer, the first time it was used. And it was a complicated thing, about seven or eight poles and seven or eight zeroes in it. and if you form a company and the company is young, very small, it's making a very small number of things, and it pays an enormous price for each component. The people who buy 100,000 capacitors of a given value get a very low price. So it's a very hard thing when you're starting out. You have to compete with somebody that is much larger than you, buying the parts at 30% or 40% or 50% less than you buy them.

So what we did is we took the equalizer. It was a much more complicated one, actually, than this. But we looked at it, and it turns out that if you look in the price books-- and in those days, you had to buy out of the normal wholesale books that you look at-- and you looked and you found out at that time that there was a big jump in price when the capacitors got larger than 0.047. When they went to 0.05, the price jumped up. It was just the way it was. And there were a lot of capacitors in this thing that were bigger than that.

Well, by this method that I'm just going to tell you how to develop it If you need it, we were able to go through, and if you looked at this equalizer when it actually came out on the market, if you were an engineer, you would think, God, what incredible luck these people had. All the values were 0.047. This huge number of the 0.047's. So there were at least 10 of them in there. And so all of a sudden that gave us 10 times the purchasing, the quantity, and we could get better prices. So there's a very good reason why you might want to do that. Another reason is that you might want to see if you couldn't reduce the number of elements. And I'll just tell you where it comes from, in case you ever get to this place. You can look at it.

So these are a set of equations that you can use Cramer's Rule for. In other words, if you want to solve for E2, all you do is you dump this column in here, which is IS instead of the Es here. IS 0 0. And you use Cramer's Rule on this. Cramer's Rule, you get a co-factor over a determinant.

Now this determinant here is made up of these coefficients. It belongs to the network that you have. Now if you multiply the rows and columns, you have to do it with symmetry. Otherwise what connects to this node and this isn't the same, depending on which way you looked at it. That's a disaster. You saw that that thing has to remain symmetrical. So you have the freedom to take and multiply a row and a column by some factor, A. But as long as you choose a row and a column that still exist in this matrix, this determinant up here, because then when you multiply by A, the whole determinant goes up by A squared. And so does this one. And the ratio of these determinants is what you want. It's the transfer function you want, and it's unaffected.

And so what you find out-- it's very interesting-- in order to maintain symmetry, the diagonal term will be multiplied by A squared, and the off-diagonal terms will be multiplied by A. And that gives you freedom to play with these elements, change their values, change whatever you want as long as you make sure that you pick a row and a column that the co-factor still has. And change elements, change topology, sometimes. It's a big advantage. So I just tell you that. It falls directly out of Cramer's Rule, if you ever need it.

Any questions about this business? Yes.

MALE SPEAKER: Can this be extended to work with transformers or dependent sources also? Or doe sit only work with one-port networks?

AMAR BOSE: No, it works with multi-port networks, because getting a transfer function from here to here, to any other port that I want to define-- but as to working with dependent sources, I am not sure at the moment. I think the answer is no. And transformers, I don't know how to do it with transformers. Dependent sources, I can-- no, that will destroy the symmetry of the determinant. The answers no to both of them. At the moment, it's something you can use for passive networks, but the two examples I've given you really do come in handy sometimes.

Now if I asked you at this stage to write another network, I just want to know what you would do. Not do it. Suppose I asked you to make another network in which I came rushing in here and I changed this to i1 and this to i2, et cetera. In other words, these became loop equations. What would you do?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: That's the easiest, take the dual. But you could go back and you could say, OK, this is the sum of the elements that are around the loop that's loop one. These border between loop one and loop two. These elements are impedances now, and they border between loop one and loop three. And you could make it. But otherwise, what you could simply do is you could go over here and you could put a dot here, a dot here, a dot here, a dot here, a dot here. A heck of a lot of loops, and a dot outside for the datum node. One, two, three, four, five, six. A seven-loop network will be the dual of this. And you just put the lines here, and a resistance of value 1 will connect between this node, whatever you label this-- E6 or whatever-- and the datum. And an inductance will connect between those nodes, et cetera.

OK, any other questions on this? OK, now I'd like to give you a mechanical system that's about as complicated as we will need to go, and see if you can write an electrical model for me. Let's have some sort of a big bell jar here. And this has, let's say, a mechanical mass one. It has a handle up here which we can connect to a source, or we can shake it or whatever you want. We have a spring. This is mechanical stuff, so we have a spring here. Let's say we have a mass here, a spring here, mass here. And we have some plates. This is a cross-section. If I look down on this thing, it's a circle. It's a jar.

So let's just use this to indicate a viscous friction, r mechanical one. Another viscous friction, r mechanical two. This is a compliance, C mechanical one, C mechanical two, mass two, mass three. And let's just make an electrical network that will represent this mechanical system, so we could get the frequency response, let's say, if we wanted to, from the handle up here to the movement of the mass here.

Now all the masses and everything are moving in one direction. What the mechanical engineers have to face sometimes is things a little more complicated than we do, where you have a box full of stuff and you kick it and things go that way and they go this way, and then you have to keep track separately of what's going on the different axes. But here everything's going up and down.

OK, so I want that network. Here are the input terminals here. And I don't have any source shown here. I could put a velocity source or current source or whatever. A force source. And now whenever we say we want an electrical network for a mechanical system, the first thing we have to do is specify, as we choose, what's analogous to what. So let's take, in this example, voltage analogous to velocity, and therefore, of course, current will be analogous-- let's see. No. Let me just see now. The across variables I'm going to take as voltage, the through variables as force.

OK. So here are the two terminals on the network. This'll be the datum node. And now I want to fill in the rest of it. Or you want to fill in the rest of it. Come on, now. I haven't done that bad a job up until now. You should be able to tell me what goes in there. What connects to this terminal?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Capacitor, representing what?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: An M1, capacitor two, of value MM1. Now don't try to remember that a mass goes into a capacitor. All you do is write down Newton's law, and you make force analogous to current. And immediately the equation shows you that a mass is a capacitor. And therefore you know that the inductor is going to be a spring. OK, so that takes care of this fellow. That means that even if there were no insides here at all, you're having to shake a mass relative to ground. So that's MM1, mechanical mass one. Anything else connecting to that terminal? What?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Inductor to?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: To?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: To another node. Yeah. Here's another node here, that's obviously something that has a velocity that's different than there. So let's put a node down here. OK. So what do you want from this node? Capacitor to ground. Value? MM2. This fellow right here, because he's moving with the velocity of that node. Again, when you put things in series, remember, whatever your through variable, the through variable's got to be the same. Whenever they're in parallel, the across variable's got to be the same. It's that simple.

OK, we're getting there now. What about the other spring? That goes from this node to another node that can move relative to it. This is CM1. This is CM2. It comes across, and we're down here to this node here. Now let's see. We have all the masses-- one, two, three. Oh, we don't have all the masses. There's a mass to this node. MM3. We have masses in the spring. So all we have to account for are the two viscous frictions. OK, where would you like to put the first viscous friction in the network?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Parallel to CM1. The reason is it goes between-- this is RM1. I think little r the one we-- it doesn't matter. But you know that this r is, in an electrical network, a voltage divided by a current, because the across variables are voltages. It goes between here and here, because one end of the viscous friction is this mass, which is here. So it must connect there. And the other end of the viscous friction is this mass, which must connect here. And finally, where do we put RM2?

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: MM1 and MM3, exactly. That's where the two ends are. You just find out where the two ends of the dashpot are, and that is RM2. So now the frequency response of this network-- let's say, from here to here-- this would be U3, the across variable. This would be U2. So U3 would be the velocity of this fellow. U2 would be the velocity of this fellow. And you can find whatever you want. If you put an excitation on here, whatever you want to choose, that will get this network going. You could find the voltages, which are which are numerically equal to velocities. And by dividing by the impedance, you can find the force on MM2. You can find the force through the mechanical circuit, whatever. You'll have some interesting problems on this, that I think when you came into the subject would look very, very complicated to electrical engineers, but may not to mechanical engineers. You can see how you can carry your disciplines around and get the frequency response of this once you have that network. Any questions on this? Yes.

MALE SPEAKER: If you wanted to instead use the other analogy and say that voltage was current, would that simply be the dual of that circuit?

AMAR BOSE: Yes, exactly.

MALE SPEAKER: Now, when you do that-- earlier in the lectures, you said that mass is always taking reference to ground, because that's where the other terminal is. But when you take the dual of that circuit, [INAUDIBLE]

AMAR BOSE: Yes, then your electrical circuit will have masses floating around any way they want. Thank God we have the freedom to do that.

OK. Let's see now. This was sort of unscheduled, but-- oh, no, mechanical circuits. That's good. Let's see what we do now. Acoustical elements. Aha. What we're going to do is derive the acoustic elements, lumped parameter elements, two different ways. And there's a reason for deriving them two ways. The two different ways will tell us something about the element that wasn't so obvious from the other way of looking at it.

The fundamental way for us, we'll start with tubes, which you've all done before, but it doesn't hurt to look at again, in the end because they're going to be lumped parameter models. The tube is going to be all very small compared to 1/6 sixth of a wavelength. Its length here is going to be small compared to that. And I'm going to be interested in the impedance looking in here for this element. And we're going to find out that when this is small compared to a wavelength, to be sure, this looks like a lumped parameter element, which we've already seen. I did it for the closed one. I'm going to do it again. And I think you did it for the open one. We'll do it again.

First of all, let's just go back and get our equations for this. The transmission line equations tell the whole story. P is P plus e to the minus j omega x over c-- x is this axis-- plus P minus e to the plus j omega x over c. U is P plus over z0, e to the minus j omega x over c, minus P minus over z0, e to the plus j omega x over c. The boundary condition here on an open tube we say is 0. Very shortly-- not today but coming up pretty soon-- we will find out that it's actually not 0. But it's very, very small, the impedance looking out here. We say the pressure is 0. That's equivalent to saying, impedance is the ratio of pressure to velocity. So that's like saying the impedance is 0, looking out into here.

But you know already, from energy, it can't be 0, because the if a wave came down here, some of it's going to get out. And that's going to radiate, and so there must be some little component of pressure that's out here. And there is. But what really counts is-- you remember when we looked at this vector diagram with the 1 plus gamma and all-- if the amount going out is small enough, you get, essentially, a total reflection here. As far as what's going on in this tube, if you let a little bit out here, the reflected wave will have a magnitude that is extremely close to the incident wave, and therefore, for all your conditions in here, it will be essentially equivalent, if you made the assumption the pressure was 0.

Now to show you that pressure really is so small out there, we'll do that. But that'll be radiation resistance. So for here, the boundary condition is P of 0, and 0 and omega, is identically equal to 0 for an open tube. So when we put that in here, we get this implies that, if that's 0 at the origin where the whole exponents become 1, then that implies P plus is equal to minus P minus. So what goes back is a wave of negative amplitude, so that at this point here, they cancel.

So if you put that in here, you see right away-- let's look now for the impedance here. I think we can look right away, because you've done that a few times. z is equal to P minus. I told you I'd drop the business of x and omega here. It's quite clear from this expression that that P is a function of x and omega. So it's P-- just to remind you here I'll put it in-- P of minus L and omega. Just this once. U of minus L and omega. That's the z I'm looking at here.

OK. So I put P plus is equal to P minus. I factor it out. I get P plus e to the-- and I'll reverse these so you know what they look like. With a minus sign, P plus is equal to minus P minus. Hang on. Watch out when I do this, because I may need help. e to the minus j omega x over c. Downstairs here, P plus over z0, e to the plus-- I'm going to put this one first-- j omega x over c plus e to the minus j omega x over c.

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Oh, I already did put in minus L. It's easier to do this than it is to change all the other ones. OK. We'll change it to L afterwards. OK, that looks good. Now this thing looks almost like a sine except for a 2j under it. Except for a two under it, this is a cosine. So z is then equal to--

MALE SPEAKER: [INAUDIBLE] plus j should be minus j now if you're not going to put [INAUDIBLE]

AMAR BOSE: No, it works out OK there, because--

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: I put a minus out here. OK? Thanks.

OK. P plus. Now whenever you're looking for impedance, for God's sake, the complex amplitude had better disappear. Impedance is never a function of the source. It's only a function of what you're looking into. So they do disappear. So we have minus z0. Now here's where I have to watch out. This is a sine, except that I need a j downstairs. So I put a j upstairs. So I have a j. The 2's I would park below this and below that. They go away. So this is a sine upstairs and a cosine downstairs. And it looks like sine over tangent over c.

Now this, I can take away the minus sign. It's jz0 10 omega L over c, because L is minus x. That wipes out this. So I have this. Now for L much, much less than lambda over 2 pi-- and I'll let you see where that comes from. Omega over c is 2 pi over lambda. And if this argument is small compared to 1, this tangent is approximately the argument. So for this condition, this is approximately jz0 omega L over c. That's where the lumped parameter comes under this condition.

OK. Now z0 is rho 0 c, so I could write this as j rho 0 c, so the c's go out. j omega rho 0 L.

Now this impedance is a pressure over a velocity. When we go to acoustic circuits, the convenient variable-- it will become very apparent-- will be volume velocity. Pressure over velocity, complex amplitudes, is called specific acoustic impedance. Pressure over a volume velocity, complex amplitudes, is called acoustic impedance. I don't care if you remember the names or not, but that's what they're called.

So this was a P over a U. I don't want to tag along all the time some sub-note on this to tell you whether it's acoustic impedance or specific acoustic. So I'll just write one thing like this. P over volume velocity is the new z. I don't know what you want to call it. Maybe I'll call it that for the moment, just to keep your notes straight. And then pressure over volume velocity-- well, pressure over velocity divided by the area, A of this thing, is pressure over volume velocity. P over U times A is P over volume velocity. So this is pressure over volume velocity. I have to take this, divide by A. It's j omega rho 0 L over A.

So if I use pressure as analogous to voltage, which I will put as a cap V-- God, I have too many caps. Well, I have volume velocity denoted. And volume velocity therefore is analogous to current. If I use that analogy, what is that element?

MALE SPEAKER: Inductor.

AMAR BOSE: It's an inductor. j omega something. So where you have here pressure is analogous to voltage, volume velocity is analogous here, and the value of this inductor is rho 0 L over A.

Now looking into this tube such that you're looking at low frequencies, where the wavelength is very large compared to the device, you see an inductor in this analogy. Now I want to do this another way, and we'll get some insight out of that. Remember, another way of expressing this thing, the tangent of the function, was the sine omega x over c, over cosine omega x over c. That's whatever the constants are. z, z0, or jz0.

What happens here is this open tube, if you plot it-- as you, I think, have done on your homework-- if you plotted the standing waves, you'd see the following thing. Out here at x equals 0, you have a pressure which is 0. However, it's a sine wave, very, very, very-- it goes up and goes down. It's a big sine wave, because the distance L might be a very small distance here. And you have a velocity that's looking like this. And it starts down. So if you're down in here, if L is in here somewhere, you have essentially a constant velocity over that whole length of tube, length L. And the wavelength is very large compared to the tube. And you have a pressure which of course you knew-- it gets to 0 there, we set it up that way, and it rises a tiny bit by the other end.

So what this tells you is essentially this thing has no compression. It's moving like a mass, because for all practical purposes, first-order velocity here is the same as velocity there. The slope of this cosine is 0. Yes, there is a little bit more pressure at this end, as you could see from that line coming up, than there is at this end. And you're effectively pushing a mass.

So let's see what Newton has to say about that. If we had a tube of length L, cross-section A, and we pushed on this end of it, back and forth, we would have force, which is pressure times area is equal to M d dt of U. I'm taking little liberties here that maybe I shouldn't. I don't know if you're-- can I do this? See if it makes any sense to you. Then I can write it all in terms of complex amplitudes.

Newton had to say force is equal to M d dt of U. If I let little f equal cap F e to the j omega t, and of course, little u equals cap U e to the j omega t, the e to the j omega t's go out, and I get exactly this relationship. So differentiation, as we know, in the time domain drops out of j omega in the frequency domain. If you have any problem getting from this expression to this, either ask me now or ask the teaching assistants. But that you have to be as familiar with as your name. It's time to frequency domain and the simple operations.

OK. So let's see. What did we want over here? We wanted pressure over volume velocity. Well, we could write that as Mj omega. U is equal to volume velocity over area. So OK, let's see. Pressure over [INAUDIBLE] velocity. OK. Pressure times area is the force. Now I get pressure over volume velocity is equal to j omega over A squared, at the moment, times M. But what's M? j omega a squared. M is-- where's the tube? Here. It has a density rho. And so the total mass then is rho 0 times the volume, which is A times L. So that kills one of the As, and we get j omega rho 0 L over A, which is exactly, I hope, what this one is. j omega rho 0 L over A.

So you could have looked at this thing ahead of time and sort of said, well, I think the tube is behaving like this. But you have no way of knowing that it is. You go back to the waves, and the waves tell you that if this thing is short enough, the velocity is essentially constant. If the velocity here is a constant, then the whole air in there is just moving like this. And so Newton's law gives you exactly the approximations that you would derive from that one.

Now an even more interesting result comes when we try to do this for a closed tube. I will start over. I'm going to close the tube. Transmission line equations, again, are as written here. Now the boundary condition changes. Closed tube, it's U of 0, and omega is identically equal to 0. And that implies that P plus equals P minus, because exponents are 1 at x equals 0. If P plus equals P minus, then U is 0 at x equals 0. So that gives us an impedance. If I first define the impedance as P over U, that gives us that P's go out, the z0 comes upstairs. Oh, boy. Now we have a cosine upstairs and a sine downstairs. z0. The cosine omega x over c upstairs. And this is going to turn into a sine omega x over c.

Now to get this fellow to a sine, I had to put a minus in front of him. And I had to put a 2j down here. So a minus and a 2j gave a minus j up here, I hope. So that gave us a minus z0 over j cotan omega x over c. Remember, cotan expansion is 1 over the argument minus 1/3 the argument cubed, et cetera. It's essentially equal to 1 over the argument when this quantity is small compared to 1. So z is approximately equal to minus z0 over j omega x over c. But x is negative. x less than 0. So I'll put in L equals minus x, because I'm looking at the end there. L equals minus x. So we have a minus sign that wipes that out. We get z0 c over omega L over j omega L.

Now, z0 is rho 0 c, so this becomes 1 over j omega rho 0 c. So that's L over rho 0 c squared. So in terms of specific acoustic impedances, which are a ratio of pressure to velocity, this element, when the tube is small enough and you look in here, it is given by-- if the other one was an inductor, this is obviously a capacitor, when the across variable is the same analogy that we used here, pressure analogous to voltage and volume velocity analogous to current. Now I have to get that thing in the shape of pressure over volume velocity.

Pressure over volume velocity. This is what we're always going to use in acoustics. I'll just let that be some symbol here for the moment, just to tell you it's not the same z. But the z is defined by this equation. OK. Is equal to pressure over velocity divided by A. So that's volume velocity. So that is equal to 1 over j omega A times L over rho 0 c squared. But A times L is the volume, V0, of this whole thing here. V0, the volume of the tube. So this is 1 over j omega V0 over rho 0 c squared. And now that element is a capacitor in the analogy on the board. Pressure analogous to voltage. It's a capacitor of value would V0 over rho 0 c squared, where the across variable is pressure, and the through variable is volume velocity. Or when you make the analogy to electrical, this is voltage and this is current.

OK. Now very interesting, let's do this a different way and we'll find out something totally different. We'll find out a different insight that comes from it. But we still have to go back first and look at the expression, namely the impedance here. The pressure is a cosine wave, the envelope of it. The velocity is a sine wave. Velocity was 0, so if this wavelength is very large, this velocity is sort of increasing a tiny, tiny, tiny bit, and it goes up way over there. The pressure on the other hand is constant, constant, constant and it's slowly going down. So this is P, and this fellow is U or volume velocity, depending on whether you put area in there. So what it says is when you are close enough to the end, the pressure is constant everywhere.

Now if I think about that, it doesn't make any difference what shape that thing is. In other words, this could have been a shape like this. It could have been a big tank, irregular as heck, with an opening here. But if the whole dimensions are small compared to the wavelength, it doesn't matter where waves go in here. They hit a hard surface, they reflect back. The wavelengths are very, very large compared to the dimensions of the thing, and the pressure is constant throughout here. Well, if that's so, we need to get a relationship. If I want the impedance in here, all I need to do is get a relationship between pressure and velocity here. Where would you go for that?

MALE SPEAKER: Gas law?

AMAR BOSE: Gas law. That doesn't quite relate pressure straight off to velocity. What did it do? It related pressure and volume. But when you make a change in volume, you make a small displacement. Well, the rate of making the displacement is velocity. And all you've got to do is get the pressure change in there.

So what we'll do now is consider a strange shape that comes down to some opening. And we imagine that we maybe have a piston there or something, and we're looking in here for that impedance. And this thing has a volume, V0. So we would have, from the gas law, P total V total to the gamma is a constant. We can go through these steps pretty quickly, I think. dPt is just dP, taking the differential of both sides. V total to the gamma plus-- it's a product of two-- P total gamma V total to the gamma minus 1 is equal to a constant.

MALE SPEAKER: [INAUDIBLE]

AMAR BOSE: Yeah, thank you. dV total, which-- that's d tau. I've always used it. I'll still use it as d tau, as a constant. OK. Let's get some of this stuff straightened out. We can divide through by the V. We'll get dP-- maybe I'll take it over to the other side, in fact-- minus P total gamma. This goes down underneath. So it's V total d tau. I think that's got it. Yeah.

OK. Now dP-- I want to get this thing related to volume. d tau, change in volume, is A dx in this case. I'll put the x-axis right along here. And this area A is the area of this piston looking in there. And x is along this axis here. So d tau is minus A dx, because if x is positive this way when I give it a shove in the piston, the volume gets smaller. So that gives us, then, dP is equal to plus P total gamma. Oh, in fact, let me get rid of some of this stuff. Multiplying the whole side is P total, but remember, P total was the tiny, tiny thing, 10 to the minus fifth, less than anything that we're dealing with due to the sound wave. And P0 was the big one. So this is good enough to be P0. Same with the volume, V0. I put a plus sign here because I'm going to put in A dx here.

OK. Well, we don't quite have it yet. If we take this thing and put dt under both ends, we have the dt dx, and that gives us velocity, which is what we want. dP dt is equal to P0 over V0 A dx dt, which is u. Oh, is there a gamma missing? Yeah, thank you. Gamma.

OK. Now I want to go from this to complex amplitudes right away. If I let this little p, real time, equal cap P e to the j omega t-- differentiate it. It drops the j omega out. So whenever you see dP dt, if you want to go to complex amplitudes, it's just j omega P, is the complex amplitude of the pressure due to the sound wave, is equal to gamma P0-- atmospheric. V0, the volume, or whatever I let that be. I don't know what symbol I used. Well, Vt. A and U. There's no differentiation there.

Now if I wanted to get it in terms of volume velocity, it's gamma P0 over V0. A times volume velocity. Oh, I already have it. A times velocity is volume velocity, so VV. And finally, P over VV is equal to gamma P0 over V0 times-- that's all. Oh, yeah. Over j omega. Now gamma P0 is rho 0 c squared. Rho 0 c squared V0-- volume-- j omega, equals 1 over j omega V0 over rho 0 c squared. And that should be exactly what we got up here.