Amar G. Bose: 6.312 Lecture 10

AMAR BOSE: I saw the results of the quiz, and you did very, very well on the quiz. I don't know if you know that yet. But the results were excellent. Congratulations. We missed on the quiz, in the sense that it turned out to be too long. Hopefully we'll learn to correct that by the second one. But in spite of that, you did very well.

Last time, we spent a lot of time-- not anticipated-- talking about Kirchhoff laws for electric circuits, and showing that they came from Maxwell's equations and that they really aren't true, but that the sum of the voltages around a closed path are not equal to 0, but they're really equal to the rate of change of total flux through the loop. Now the flux through the loop, of course, gets bigger and bigger as the loop gets bigger. And of course, as the frequency gets bigger, your d phi dt gets larger.

So the error in Kirchhoff laws, if you wish, gets larger with higher frequency and with larger loops. You see this in the 60 hertz region, even when you make a sloppy circuit with long wires. There are a lot of 60-hertz fields running around here just because of the power lines, and you'll get all sorts of hum in your system from that. But when you are designing other systems, even in audio equipment, it's not at all hard to pick up radio stations in pre-amplifiers. So you should realize where they're coming from and be able then to minimize that.

We, I think, didn't mention that in the mechanical side, there are also the Kirchhoff laws if you wish. And we don't have to, again, but we chose to think of the through variable as force. And if you had mechanical linkages coming in here, let's say like that, force one, fourth two, force three, your Kirchhoff's current law, the KCL, if you wish, for the mechanical would read summation of forces over the whole node equal to 0, which in this particular case would read summation of forces in or out 0, which would read in this case, f1-- if I do it in-- minus f2 plus f3 is equal to 0. That's why, with the bias of an electrical engineer, we like to sometimes look at a mechanical circuit and say that force is the through variable.

I had two of our mechanical engineers come up afterwards and say, how come you were looking at it that way? We think of force as the across variable. Well, it turns out whichever way you think of it, you develop insight into the circuits that way. And that's fine. And then the other way will seem a little strange. We will have to use both, because of something that isn't obvious to you now, but I've mentioned it a few times.

We're going to actually have to make one circuit that encompasses electrical, mechanical, and acoustical disciplines, and has what we call a transducer. It's like a transformer, but it changes energy from electrical to mechanical, let's say. And when we do this and try to make one circuit out of it, we're going to have to involve disciplines in which you will change around from through variables. They might be force in one situation and they might be velocity in another. And the counterpart in mechanical engineering-- if these are your linkages or your elements or whatever, the counterpart to the KVL, if we use velocity as the across variable, V1 plus V2 plus V3, would be summation around the closed path of the Uj's is equal to 0.

And in this particular case, if I took [UNINTELLIGIBLE] I've drawn it counter-clockwise. It doesn't make any difference, of course, but if I take it counter-clockwise and start here and say the summation of the drops-- meaning plus to minus-- I'd have U3 plus U2 minus U1 equal to 0. And that is for the mechanical, the KVL and KCL.

Now what I want to do is take a little detour into just dealing with electrical networks, because most of us are familiar with that, and just the writing of equations for electrical networks, and make sure that we understand that issue. And then we'll be able to go and write mechanical networks, electrical networks. And it'll all be clear, if we get just this part clear. I'm going to do this just with a couple of examples, and we'll generalize from those, which by the way is the way you would do it if you were doing it the first time. Start with the simplest examples, look at them, see what's the essence of it, and then you can generalize.

OK. Let's see. Capacitor, resistor, resistor, and an inductor. A simple network. There's a voltage source here. Now if I wanted to find, let's say-- that's the source that's driving that. It could be any kind of a source. It's just a complex amplitude. It can be an exponential. The way I've written it, it's a complex amplitude. So you'd find the complex amplitude at any point here, multiplied by e to the ST. And if it was a sum of them, if it was a transient input, it'd be the sum of the responses.

Now suppose I wanted to find the complex amplitude of the voltage across here. Or never mind complex amplitudes, even, for what I'm talking about here. I want to find the voltage, for any voltage that I would put in there. I don't have to call it even a-- let's just leave it as a voltage. We might choose to do it by complex amplitudes. We might not. How many equations would have to be satisfied for me to find, let's say, the voltage across here?

How many? Quick. No idea? Did you have networks? Two? Four? I heard two, I heard four. Three, five, two? Two twice. Any more? Come on, let's speed it up a little bit. We've got two, three, four, five. It's clearly not one. I'm going to tell you, it's more than five. Anybody want to guess at that? Hey, now this is a pretty simple network, and you did tell me you were EEs, yeah?

SPEAKER 1: Two for every component?

AMAR BOSE: Two for every component? Let's see. One, two, three, four. That would be eight. Higher, higher. Six? Well, we have six. All right.

All right. Let's take a look. There are two for every component. Somebody mentioned that. That's right. That's eight. There are eight. Let's identify them. Vi relations. In other words, there's an expression for voltage, V equals ir, remember? There's an expression that relates the voltage and the current across each element. That's one equation.

Now a source only constrains one of the two variables. The other one is an unknown. So there's one source variable. I should just say relations. I'll just say eight variables associated with the elements. And one source variable. Total, nine.

Now how many of you, if I wanted the voltage across here, I gave you a cap Vs, and I wanted it across here, this voltage, how many of you prior to coming into this class would say that you could have determined what that complex amplitude is? OK. That's more than half. How many equations would you have solved? Because you didn't tell me nine. How many? I've got more than half of you that just told me you could have done it.

SPEAKER 2: One second-order differential equation.

AMAR BOSE: One second-order differential equation. that would work except you'd have to find that equation when you had all those darn-- I mean, there's an enormous number of variables there. And I don't know how you'd find that. The answer is, you could express it that way, but I don't know how you'd get the differential equation.

SPEAKER 2: Two voltage dividers.

AMAR BOSE: Two voltage dividers. You could look at the whole thing as the impedance across in here. You could find this impedance over the sum of the impedance to find that voltage. You could find that. That works in this network. It doesn't work in more complicated ones. Let's see if we can't do it generally.

Let's see where we get all our variables from, in all our equations. We have nine variables total. We'd better have nine equations. So let's see where we would possibly get them. We get eight from the Vi relations. Do we get eight? Hey, I'm sometimes sleepy when I come in here, but-- eight equations from the Vi relations? Four. Eight variables. Four equations. V equals ir. OK, so four equations.

Now what other equations are there that have to be satisfied?

SPEAKER 3: [INAUDIBLE].

AMAR BOSE: Sure. Vi relations only give you what each element has to constrain about its voltage and its current or its across variable and its through variable. But it doesn't say anything about what happens when you connect them. So you have KCL equations. How many KCL equations would you say for this system? Two? KCL equations are written for nodes. How many nodes are there? One, two, three, four. How many KCL equations? One. KCL says, let's say, the sum of the currents going out is equal to 0, of any node. The sum of the currents going out of here is 0, out of here is 0, out of here is 0. When I add them all up, that is exactly the sum of the currents coming in here.

In other words, if I add this is a current going out here, well, when I add the currents going out here, that cancels with that. So if you have any doubt about this, write KCL for this node, this node, and this node. And you'll see that these currents going back and forth in the sum of the equations cancel out, and they all equal the current coming into here. The sum of the currents flowing out of this node, this node, this node equals this one. So KCL you can write for n minus 1 nodes. It's sufficient.

OK. KVL equations. How many do I need to write? Two. Let's see. If I say two KVL, l could have this loop here. That's a window, a closed path. I could have this one, but I could also have this one. But if I write an equation for this closed path and this closed path and add them, I get exactly the equation for this closed path, for exactly the same reason that the integral of the curl over an area is equal to the line integral of the vector around the area. Because the curl is a whole bunch of these little things like this. You add them all up, and what's going this way here goes this way next to it. They all add out, and you get the periphery. So this is just a macro look at the integral of the curl.

And if there's any problem-- this isn't obvious, it's hard for me to tell, but if this isn't clear the first time, just write the KVL equation around here, write the KVL equation around here, and add the two together. And you'll see that this element doesn't appear, and you get the KVL around the whole loop.

OK. Four and three and two. Nine equations. Now you're right, those of you who said you wouldn't write nine equations to solve this network. Now how many of you have heard of the loop method? Wait a minute. I want you to raise your hands high. This is a good measure for me, because I have to know it to know whether-- OK, down. How many have you heard of the node method. More. That's interesting. OK. Let's look at these methods for writing equations.

There is no way around it. In order to find any response in this network, you have to satisfy nine equations. And yet if you think of it, you're only going to write on the loop method two equations for this network. Let's see what's going on. You have to satisfy all of these equations.

All right. The loop network, what it does-- it tells you that, OK, you put a current around here. We'll call this i1. You put a current around here, you call it i2. i1 and i2. You just draw them in. These are called loop currents. Now a property of the loop current is, if you look at it, it's obvious-- namely, that it automatically satisfies KCL, because this current, every node that it enters, it leaves. So KCL is automatically satisfied whenever you draw a current that closes on itself in a network. So just by drawing these two currents, we have in fact satisfied KCL.

Now the thing that remains to be proved-- and it's in one of the references, [? Beau Stevens' ?] book for example-- that if you have a planar network, i.e. one which you can draw on the blackboard, not a cube with a diagonal element in it, if you can draw it on the blackboard without any crossing wires, that assigning a loop current to every window satisfies KCL. But the thing that you would need to prove, and do prove in network theory, is that any distribution of currents in this network that do satisfy KCL is representable by a set of loop currents in the meshes, or windows as they're called.

In other words, let's say it again. It's clear that assigning loop currents satisfy KCL. And I'm telling you that you can then prove, and you have it in one of the references, that any distribution of currents-- if I write down any numbers in here that satisfy KCL, I can represent that set of numbers by loop currents.

SPEAKER 4: [INAUDIBLE] non-planar circuit, the reason why you can't use meshes?

AMAR BOSE: No. For a non-planar circuit, what happens is it gets a little bit funny as to how do you identify the right number of closed loops. Now you have a three-dimensional sort of a thing here, and it's easy to identify the closed loops in a planar network, and to see that the sum of them equals any other loop that you can make within the perimeter. But when you go to a three-dimensional thing, that's not quite so obvious. But the next method that we will talk about takes care of that, too, in case. And are a mighty few, fortunately, of non-planar ones that we have to deal with.

OK, so the process, then, to solve this thing is we, one, assign loop currents, one to each window in it. That satisfies KCL. Now we write KVL around the loops. We've only satisfied KCL. We have to take care of KVL. So we write them around the loops. It is sufficient, as we just mentioned, to write them around all the loops of the meshes which you defined the currents in, because that then encompasses all the KVL equations that you can write, all the closed paths.

Now that only would take care of three plus two. These you take care of by inspection when writing the KVL. In other words, I'll do it, but write Vi by inspection. It's crazy. You insert them. I'll say insert them. Insert Vi by inspection when writing the KVL equations.

Now let's do it. I assigned the loop currents. I'm on step two. I will write KVL, the sum of the voltages, around the first window or mesh up here. The voltage drop across the first-- I'll do the sum of the voltages. The sum of the voltage drops is equal to 0, or the sum of the voltage drops is equal to the voltage rise, if you wish. It's the conventional way of putting it. So the voltage drop going in this direction, across the C, is the impedance of the c, 1 over SC times i1. Voltage drop coming down through the resistor, R1, is plus R1 times i1 minus i2, because i2 goes up through it, and i1 comes down through it. So all of that, plus the voltage drop across the source, if you wish, which is a minus VS, because it's a rise. I'll take it over to the other side-- is equal to VS.

So this is KVL for loop one. And I have written the Vi by inspection, inserted them into this thing by inspection. For loop two, it's what now? The sum of the voltage drops around the second loop is 0. Voltage drop going clockwise through the resistor is i2 minus i1 times R1. R1 times i2 minus i1 is the voltage drop across that resistor. Then voltage drop across the next one is i2 times R2. Voltage drop across the coil is i2 times SL, the impedance of the coil. SL times i2 is equal to 0.

So we now have two equations and two variables. We solve for i1 and i2. And of course if you have i1 and i2 by inspection, then you have the current through any branch. A branch is the resistor or the components. The C represents a branch. The R2 represents a branch. L represents a branch. So by getting i1 and i2, you can go right away to any branch current by inspection, using the Vi relation. You can get any branch voltage. So there are two equations and two unknowns.

Now let's rearrange these. And the reason I'm going through this so slowly is what you will see will be very useful to us in modeling. Namely, you're going to eventually-- I'm going to say this now and I'm going to say it again, hopefully, several times yet today-- when you're modeling a physical process, very often what you'll do is come up with a pile of equations. If you're able to look at that pile of equations and make a circuit from it, then you can immediately use all the insight that you've developed for your circuits, looking at it and saying, oh, this is a low-pass filter. This is a band-pass filter. Whatever the process is.

It may be a physical process that you have no intuition about at all. But if you can go from these equations to a model in a dimension in which you have insight and intuition-- the mechanical engineers mechanical, the electricals electrical-- you're way ahead. So that's why I'm spending so much time doing things like interpreting these equations.

So let's gather all this stuff so that we've gathered all the coefficients of each variable. So we'll have something times I1 plus something times I2 there, and something times I1 here plus something times I2 there. So the coefficient of I1 would be 1 over SC plus R1 times I1 minus R1 times I2 is equal to VS. I seem to have fallen into using cap I's there, complex amplitudes. So I will do it all the way through.

OK. Now that should be the rearrangement of this equation. The rearrangement of the next equation would be minus R1 times I1 plus R1 plus R2 plus SL times I2 is equal to 0. Let's see. All the coefficients of this-- I2 multiplies R1, R2, and SL.

Now there's a lot to be said about this last equation. And you can right away generalize it to a whole array in a matrix of these. First of all, you could interpret this the following way. You could say that the coefficient of this term is the contribution to this VS, if in fact it is VS, as if I2 were 0. Don't worry about the physical. This is an interpretation. Don't worry about how you get I2 to be 0 or anything else.

Just looking at these equations, this is a voltage. I times an impedance. This is a voltage. If I2 were 0 and I1 were running around in this circuit, then VS would be the current times this impedance, the capacitor plus the resistor. So that says that going around the first loop, this term is the contribution that is made by I1 to VS, as if I1 were only running by itself. This term here is the contribution to VS, as if I2 were only going by itself and I1 were 0.

Let's see what sense that would make. If I2 were going by itself and I1 were zero somehow, you'd have a voltage drop that was this way, a minus I times R1. And since there was no current here, the voltage here would be equal to the voltage here. So in fact, this is the contribution to VS, due to I2 operating alone. And this is I1. Similarly, this is the equation for loop two. This is the voltage drop around-- because there's nothing on the other side-- the voltage drop around loop two, if only I1 were operating. If only I1 one were operating, the voltage around here would be exactly that. If I2 were operating only by itself, you'd have this whole set of impedances that contributed to that.

Now look what happens with the coefficients. This is a two-element thing. It could be much, much bigger, of course. But let's just look at the diagonal terms. The diagonal terms represent-- and we can see why-- the sum of the impedances around that loop for which that equation is written. This equation was written for the first loop. The impedances around the first loop are the C and the R1. So the diagonal term is exactly the sum of the impedances around the first loop, because that makes it such that this is the contribution to the voltage drop around the first loop when only I1 is operating. So diagonal terms are some of the impedances.

Here, let's take a look at this. This is equation two. The diagonal terms here are for loop two. And R1 plus R2 plus SL. The sum of the impedances around the loop constitute the coefficient for the diagonal terms. The coefficient for the off-diagonal terms are the elements that border between the loop for which you are writing the equation and the variable that they are the coefficient of. In other words, I'm writing the equation for loop one. I'm looking at the coefficient for I2. So R1, I know, must be the element that borders between loop one and loop two. There it is.

Similarly, here, it's got to be the same. And that tells you something right away, that these off-diagonal terms for the respective loops are going to always be equal. And these are arbitrary directions that you can assign them. If you assign them all in the same direction, these will always be negative. If you assign them in different directions, that wouldn't necessarily happen. But if you assign them all in the same direction, your off-diagonal terms will always be the terms that border between the loop from which you're writing the equation and the variable for which they are the coefficient. And so you will always have this kind of a symmetry associated with it.

So if you're writing, for a network, a set of loop equations, and you come out with something different in this coefficient, whether it's a sign or whether it's another element in here that's different from that, you know you have a problem. And you can quickly look at your equations. You can write these now by inspection from the network. Any network here, you say, ah, first just the sum of the impedances around this loop, the sum of the impedances around this loop, the impedance that borders between the two loops. You're done. You don't have to think of KVL or anything else.

But don't remember the result. The result is very nice if you're going to be writing equations all day long. But where you got the result is the important thing. And you'll be able to retrieve that at any time, if you just go back to KCL and KVL. You'll be able to retrieve these results of the sum of the impedances being the diagonal terms and the mutual impedances being the ones that border between the loops.

OK. I need now to do one more example. Let's take another network. I'll take a network with a current source this time. Let's see. Take a resistor here. Take a resistor here. Take a capacitor here. Now I'll call this L. We'll still call this R1 and R2, and C. Now this network, I've arbitrarily decided to do what we call the loop method, which does exactly what we say here. It assigns loop currents, writes KVL around the loops, and inserts Vi by inspection. This one, I'm going to choose to do the node method.

Now this is actually how you would choose it, because there are only two loops in this, two independent loops. There are three nodes. If I chose the node method, which we're going to do over here, I'd have to write three equations. Here there are only two independent nodes. This one's not. The sum of the KCLs from there is equal to the KCL from here. And there are three loops. So I wouldn't normally choose to do the loop method on a network like this.

So now we're going to exactly parallel what we did over here. In the node method, what you do is you assign-- oh, boy. Too hot. Well, an E1, a node voltage, relative to some outside node which you can call your datum node, the one that isn't independent. We assign node potentials. Now just the parallelism you want to look at here. Assign node potentials, and that automatically satisfies KCL, just like assigning loop currents satisfied KCL. Node voltages satisfy KVL. One, assign node voltages.

Node voltages are not necessarily branch voltages. Branch voltages are the drops across any given branch. A node voltage is a node relative to ground. Node voltage here happens to be the branch voltage, but so be it. Put another resistor in here and it wouldn't be.

Assign node voltages that satisfy KVL. Why does it satisfy KVL? Well, if I can assign a node potential to each node, then the sum of the drops around the loop have to be 0, of course, because I have to come back to the same voltage that I started with. So if I add up all the drops around here, I'd better get 0 to get back to the same one. So that satisfies KVL.

Now the next thing we did was, for the loop method, we wrote KVL. But now we have to write KCL, because we already satisfied KVL. Two, write KCL for the nodes. And in writing the KCL for the nodes, we do exactly-- you see what's happening here. The parallelism is incredible. We put in the Vi by inspection. Insert Vi relations by inspection when writing KCL. Let's do it. I have to write KCL for this node. Some of the current, let's say, going out is equal to 0. Current going down here is E1 times that admittance, or E1 divided by the impedance. Whatever.

So E1 times 1 over SL is the current going down here. Current going out this way is E1 minus E2 divided by that resistance, or times its admittance, which is 1 over the resistance. Same thing. R1 times E1 minus E2. And the current going out here is minus IS, but I can take that over to the other side, as is usually done for sources, and put IS. So that is KCL for node one.

Now KCL for node two, some of the currents going out of here, is E2 minus E1. The voltage here, the positive voltage-- E2 minus E1 divided by R1. 1 over R1, E2 minus E1. That's the current going out there. The current going down here is plus 1 over R2-- some of the current's going out, I'm writing-- times E2. The current going down here is divided by that impedance, or multiplied by the admittance. SC times E2. It's all equal to 0.

Now that is exactly equivalent to this set of equations, equivalent meaning we've done the same thing in KCL here that we did in KVL there. Now let's do the rearrangement to get to the equations on the bottom.

OK. If I rearrange this, I get E1. I gather everything that multiplies by E1, and I get 1 over SL plus 1 over R1 times E1 minus-- for E2-- 1 over R1 E2, equals I sub S. I rearrange them. Rearrange the second one here. Now the R1 only appears here. It's a minus-- sorry. Oh. Did I screw up? Let's see. Well, I'll write it out and see where I made a mistake, if I did. I'll put the E1 term down here first, sorry. Minus 1 over R1 times E1. And then E2 times 1 over R1 plus 1 over R2 plus SC times E2. No, we're OK. No mistakes.

OK. Now all we did is we satisfied all nine equations up to here for this network. Two of them satisfied by just assigning node potentials, three of them satisfied by writing the KCL, and the rest of them satisfied by putting in the Vi relations by inspection..

Now let's look at these coefficients very carefully. Again, you can interpret this as, if E2 were 0-- don't worry how it gets to be 0 in the physical-- this admittance here would you give you the contribution. This times E1 would be equal to IS. In other words, if E2 were 0, that would pull this fellow down to ground, and E1 times current flowing through the inductor-- I mean, E1 over SL is the current going down here. E1 over R1 is the current that would go down here, because this is now ground. So this is the contribution to current as a result of node one. This is the contribution to current at node one, if E1 were 0 and E2 were non-zero. In other words, if E1 were 0 and E2 were non-zero, current would be coming back in, not going out, through that node. This is the contribution that node two makes to the current at node one. Similarly for the other terms.

Now look at the coefficient for the diagonals. This is the sum of the admittances attached to node one. The admittance of the inductance and the admittance of the resistance. Those are the only things attached to node one, the only elements. This, the diagonal term, is the sum of the admittances attached to node two. [INAUDIBLE]. This is the admittance of the term that borders between the node for which you're writing the equations and the variable for which it is a coefficient. In other words, I'm writing the thing for node one. The variable here is node two. This must be the admittance that joins nodes one and node two.

So again, you could write all this by inspection. If you saw a network like that and you knew this result, all you do is you say, ha, all right, that's E1. So I'm going to need admittances connecting to it. E2, sum of the admittances connecting to it. Diagonal terms are the admittances that join between the two nodes. And you can use that for any kind of a network. Three-dimensional or whatever you want.

Now let's look one step further. If any element occurs here that does not occur here, or if this were many dimensions, if you had E1, E2, E3, if any element occurs in the diagonal term that doesn't occur in the E3, E4, or any of them, that element is connected from that node to ground. If it were not connected from that node to ground, like the R1, it would appear in another node. So I can take a look at these equations right away and say-- without looking at the network at all-- there's an L that to go from node one to ground. There's an R that goes between node one and node two.

In other words, if I were just given that set of equations, I could call this node one and this node two. There is an L that goes to ground. Here it is. That's the admittance of an L. There is a conductance of 1 over R1, or I'll call this a conductance. G1 equal to 1 over R1. That goes between the nodes. Same here. That's all right. It's already in between node one, which is the variable, and two, which is the equation.

And let's see. These admittances connect to node two. Aha. Now only one of them was in any one of the off-diagonal terms. So I must have two elements here that go to ground. One of them must be a conductance, G2, which is 1 over R2. And one of them must be the admittance of what? A capacitor. So it must be a capacitor of value C. Oh, and what else? Up here, we forgot the source. All of that stuff must be equal to the current flowing in. And so the current flowing in was IS.

So starting with the equations, you can go back to the network. Starting with the network, you can go to the equations, once you see this interpretation. Now, that is very nice, because when you're doing a physical analysis of something that isn't the network, it's a physical process, and you want to model it and you get a set of equations like that, then you say, well, let me see. From that set of equations, I'll construct the network and, aha, now I know how all this works by inspection.

Again, most of the problems that you solve in life are going to be solved by inspection. There is pitifully little time in life to have to drag out and write equations or program computers for this stuff if you have to see the essence of it. Then if you see the essence, you can get somebody to solve the equations and program it on a computer if it's complex. But you've got to be able to see inside the problem, and you've got to be able to just say, I know what's happening, in the vast majority of problems, even if you go into engineering.

Now I did a little bit more interpretation of this than I did of the loop. I showed you that any terms that were in the diagonal and didn't appear in any of the off-diagonals had to go to ground. What about over here? Any terms that appear in the diagonal in the loop method, that don't appear in the off-diagonal must border between the loop that you're writing the equation for and the outside. You can think of the outside as another whole loop, if you want, just like you think of the datum node as a node outside.

Elements here that don't appear elsewhere go to the datum node. They go to what we would call ground. Elements here in the diagonal terms that don't appear elsewhere border between that loop for which the equation is being written and the outside. And all the off-diagonal terms border between two loops. Well, you have the same interpretation over here. You can say that all these elements border between two loops, if you just admit of an outside loop. The ones that don't appear in the off-diagonal terms border between that loop and the outside world. And all the rest of them appear in off-diagonal terms.

So again, looking at this set of equations, if you hadn't seen the network, you could right away say, aha, this network obviously is written with two loops. So I know that bordering between the two loops is an R1. I know that the sum of the impedances around the loop is this plus that. Another way I can say is, hey, this element doesn't appear anywhere else, so it has to border to the outside loop. And that's an impedance of a capacitor. Here I have R1 bordering between loop one and loop two. And now these fellows here, R2 and SL, have to border to the outside loop. And they're impedances, so that's R2 and L. So again, you can go from equations to a network or go from a network to equations.

Now I want to pay special attention here to the following, because we're going to need it a lot, and because it's generally useful. What I want to do is take this set of equations and just interchange. Forget networks for a second. Just imagine that somebody presented you with this set of equations, and you said, gee, I now know how, from this set of equations, to synthesize a network. Just imagine for a moment what would happen if you said to yourself, hm, I'd like to interchange the role of voltages and currents and see what I'll get. So I will write a new set of equations-- just equations, don't think about networks yet-- in which this is E1, this E2, this is E2, E1. And over here, instead of this being a voltage source, it'll be a current source, but numerically equal to the voltage source.

OK, so let's just do it. Oh, boy, that's going to be a good spread for me to see there. 1 over SC plus R1 times-- that's I1. I want to make it in an E1. I'm just doing it for the heck of it. Minus R1 times E2, is equal to IS. And I'll just keep track over here. IS is going to be equal to VS. I'm going to interchange the roles of all voltages and currents in this set of equations. No motivation, just doing it for the hell of it. OK. Now-- minus R1 times E1 plus R1 plus R2 plus SL. That equation times E2 is equal to 0.

So I have a set of equations. But now we know how to synthesize networks that behave like those equations. Let's synthesize this network. OK, let's see. All these things have to be admittances now. They were impedances over there, right? Something times I gave a voltage. We were making voltage drops. Here, something times E, whatever it is times E, had better be an admittance, and then it becomes a current, the product.

So what does this say? It says that these are now a set of node equations. So let me look at them and say, aha, node equations. Between node one and node two, there must be this fellow. An admittance, G1, numerically equal to R1. This whole term is the sum of the admittances from node one to ground, if E2 were 0. Or said in another way, anything that appears here that doesn't appear anywhere else goes to ground. Now that's the admittance of something going to the ground. What do you think that element is? An inductor, right. The admittance of an inductor is 1 over S times something. And what is the value of that inductor? C. It's an inductor who is numerically equal to C, just like this admittance was equal to R1.

SPEAKER 5: [INAUDIBLE].

AMAR BOSE: Where? This one here? No. I just copied the set of equations from there. So this is the right set of equations. Now this is an admittance here. Oh, I know what the problem is. This is an admittance which is 1 over--

SPEAKER 5: [INAUDIBLE].

AMAR BOSE: This is an admittance which is numerically equal to R1. It's not a resistor. I look at this thing. This R1 is an admittance, or in this case what you'd call a conductance. It is something who is R1 mhos, 1 over the ohms. So it's conductance. Jesus, now I'm screwed up. Wait a minute. Yeah, this is an admittance. If I only write it this way, you will interpret it as a resistor of three ohms. What it really is, is a resistor whose conductance is three mhos, or whose resistance is 1 over the R1.

How do I write this? I guess I have to write it this way. It's a conductance. If this was three ohms, this is three mhos. This is a conductance equal to-- yeah.

SPEAKER 5: [INAUDIBLE].

AMAR BOSE: Yeah. That might do it. R1 mhos. Good. Yeah, thank you. Now let's see. All right. So we got these two terms here. This is E1. Now what happens over here? Second node, E2, is here. Well, this one's already taken care of here. Now we have two more of that didn't appear here. So we have another one of these strange fellows, which is a conductance, G2, numerically equal to R2 mhos, the same as that. And since these are admittances, OK, they're in parallel. So we have an admittance of SL down here. What is that? The impedance of a capacitor is 1 over SC, so this is an admittance. And the admittance has to be a capacitor. And the value of this capacitor? L. So I'm going to leave that just as C and that as L.

Now let's see. We forgot to put in the source, as usual. The source fed current into node one. So now it's a current, of course. It's a current source, because I'm writing a sum of currents in this. Current source, IS, numerically equal to VS. If the voltage in that was three volts, this current is three amperes.

Now isn't that a coincidence. There's a certain similarity here. And if you look at the two, the inductance here just became numerically equal to the capacitance in the other circuit, whatever it was, up there. The capacitance here became numerically equal to the inductor in the other circuit. The admittances here became numerically equal to the resistances in the other circuit.

Now in this circuit, we have exchanged every voltage for every current. That's how we got this set of equations from the loop equations. We wrote a set of node equations from the loop equations. So each branch that the voltage across here goes into the current through the corresponding element over there, and vice versa. So you look at this and you say, oh, my God, we have made a network here in which voltages have been exchanged for currents. And when we did that, we found that Ls went into Cs, Cs went into Ls, resistors went into conductance, and voltage sources went into parent sources.

Now if I were to do exactly the same thing, again, of course, and just interchange voltages and currents, I'd go back to that set of things and I'd realize the first network. This is called the dual, D-U-A-L, of the first network. Now before you see where it will be necessary for us, you might wonder, big deal. Why do you do this?

Probably the biggest exercise that the world has ever seen in taking duals took place in the '50s at Bell Telephone Laboratories, according to my information. They had a number of people employed taking duals of circuits all day long. And why? Because the bipolar transistor looked like the dual of the vacuum tube model. And therefore the circuits that you would connect to it to make the different functions that you used to make with vacuum tubes would be the dual. If you had something here that was the dual of a model, you'd connect the dual of the other fellow's circuit that he designed, and now you have a functioning circuit which does exactly the same thing.

And so they were busy taking duals of all these important vacuum tube circuits and patenting them, so that they would have a big collection of patents that would govern some of the cleverest developments that had been made in vacuum tube circuitry. So there are commercial reasons, once in awhile, to take duals also, besides engineering ones.

Now let's see. Before I go any further on this, let's just first see if you can formulate a question. Gee, did I do that badly? Yeah?

SPEAKER 6: This is a little off-track, but if I took, then, my node circuit and went about and formed the dual, not that you would solve things this way, but then that would allow you to identify the independent meshes, the equivalent noded mesh circuit [INAUDIBLE].

AMAR BOSE: You wouldn't use it to identify meshes, because if you had the circuit, originally, you could identify them. Yeah. You could. Yeah. Let's see. OK.

Now if you take a closer look at this, there turns out to be a very simple way of doing the duals, which is based on everything we've talked about today, and it's so easy when you see what happened here. Namely, we said that every element in the loop equations which was in the diagonals that never appeared in any off-diagonal terms bordered to the outside loop. And in the node equations, every element that was in here that was not anywhere else on that same line connected from that node to the ground. All the other ones border between the two nodes that are appropriate.

So now let's see if I can just take that network up there-- I'll do it first, and maybe when I'm actually doing it, you'll see why, and then we'll talk about it afterwards. I'll put a dot in the middle of each loop, and a dot outside. I'm just going to do this mechanically, but I think you will have enough already that you'll see it unfold. And I'm going to call those things the nodes. This the datum node. This is the outside, if you wish. I put a dot in each loop and one in the outside loop. And I put a dot here. I'm going to call this node one and node two. And I'm going to draw lines that join these different nodes going through one and only one branch at a time.

And the element that was cut by this line-- this is node one and this is the datum. That goes here, and it becomes a conductance, G1, who is numerically equal to E1 mhos.

SPEAKER 7: R1.

AMAR BOSE: Oh, yeah. I thought I started out with three erasers. This is a disaster, by the way. If you drop anything down here, there's no way out.

Let's see. Equal to R1. Now I draw another one through here, and it's one and only one. Connect to these two. Well, that's going to the outer loop. So that must go to ground. And that capacitor becomes an L who's numerically equal to C. I draw another one and I connect each node connected through one and only one element. So VS, that becomes a current source now, numerically equal to VS. By the way, the arrow dominates. This is the value. The arrow says it's a current source. V sub S says it's a current source of value VS. If I had this, that is a voltage source of value IS, numerically.

OK. So now, let's see, I have a couple more left. I go through there, I'll put that one down here, and that's G2, again numerically equal to R2. And I have one more I can draw. And that is going through an inductor to the outer loop, which means here to the datum node. And that is a capacitor numerically equal to L. The reason you don't have to do all this monkey business for the capacitor and the inductor, of course, is because they say what they are. This thing says it's an inductor, this thing says it's a capacitor, and then this says it's the value. But unfortunately, the symbol for the conductance and the resistance is the same.

Now that of course gives me exactly this dual. But the whole thing is based upon the fact that in the loop method, what bordered between two loops connects between two nodes over here. So if you put the node and the node here and you run a line through whatever bordered between these two loops, that must be the element that connects over here. Whatever bordered between this loop and the outside world, we saw in the dual connected from the node to the datum node. So when I go through here between these two nodes, I encounter the capacitor. So that becomes an inductor that connects to the ground. When I go from here, similarly, it becomes a conductance. This becomes a capacitor.