Amar G. Bose: 6.312 Lecture 05

DR. AMAR G. BOSE: But it's a fun item, and I hope we get time to get to it. Last time, we developed the so-called transmission line type equations, which express everything that can be expressed in terms of the complex amplitudes of signals traveling in one dimension, plane waves in one dimension. The positive going wave, the negative going wave, and for pressure, and then also for velocity.

Then we said, let's see what we can begin to get out of them. What comes out of these equations is amazing, and a lot of our insight will be developed just from there. And these, as you remember, fell right out of the solution to the wave equation, the general solution, P of x, s, e to the st, and broke that down into a positive going wave and a negative going wave.

When we got this, we said, all right. Let's look at a one-dimensional wave, plane wave, which we said we can also consider as a wave in a tube. Or you can consider it as an infinite wall moving along, it's initiating a wave going down that direction. And we said that the wave that we've initiated is a P-plus, going this direction. And we just assigned this to be the initiated sine wave.

Then we wanted to find out what the pressure and the velocity at any instant in time at any point in space, clearly to the left of the origin, what pressure and velocity were. And we got the following expressions right out of the transmission line equations. How did we do that? Well, we first got P of x, s from them and U of x, s.

How did we get that? Well from this, we expressed this as a complex exponential of the form P of x, s e to the st, which you've done already in your homework. Plug the complex amplitude in for P-plus, and a boundary condition gave us the relationship between P-plus and P-minus, this boundary at this point, velocity had to be 0. And so all we needed was the P-plus of s to get these, P of x, s and U of x, s. Once we had those, you multiply by the e to the st, s was J omega, and you got the time domain answers.

Now, I think when we finished last time, we were talking about the picture of these two equations. And I talked about an envelope. Now, let's go a little further here. Take the white one-- what does the envelope mean? This is for pressure. This quantity here you can look at, once you fix a point in x, as the amplitude of the cosine wave that exists in here. In other words, take a point, any point you want, and, in x, put a pressure microphone in this tube at that point, connect the pressure microphone to an oscilloscope, and what will you see? You'll see on the oscilloscope a cosine function given exactly by this whose amplitude is given by this.

And this white curve is the envelope of that amplitude as a function of x as you go away. The actual amplitude is cosine omega x/c, which is this here. But if you draw the other side of it, this is the amplitude of the sine wave that you would see going up and down.

Now, you have to be a little careful in one sense. It's very interesting that the phase of whatever you would see here on the osciloscope is governed totally by this expression. And the amplitude by this.

But watch out, because the cosine changes sign at this point. So though I've drawn an amplitude, I have to realize that when I go through this point, the cosine changes sign. And so what I would be reading right to the right of this point would be a minus quantity times this, like 180 degrees phase shift. And similarly, when I get to this point, the cosine would change.

Similarly, for velocity, when I got to this point, the sine function would change sign. So it's not just an amplitude. You can think of it as something that flips through 180 degrees in phase.

Now, if you were to take a snapshot in time of the whole thing and see what the pressure was at that one instant of time all the way along here-- I've only drawn the envelope, this part of the amplitude. Now, suppose I came up instead with a camera and said, aha, I'm going to catch at that instant of time what the pressure waveform looks like. What do you think might happen?

Well, if I happen to take an instant of time in which this cosine was 1, what I would see is exactly the curve I'm tracing out now. If I flashed the camera at an instant in which the cosine term was minus 1, I'd get exactly this curve. If I flashed the camera at an instant in which the cosine was 0, argument pi over 2, guess what I'd get. I'd see a 0 all the way down at that instant of time throughout. If I flashed it at some other time, I might find a curve that looked like this. Whatever this amplitude was-- yes?

SPEAKER 1: The second terms [INAUDIBLE PHRASE].

DR. AMAR G. BOSE: Second terms are-- whoa, oh my god. Thank you, thank you, thank you. Omega-- were they last time, by the way?

SPEAKER 1: [INAUDIBLE].

DR. AMAR G. BOSE: Thank you. Whew. Omega t plus 5. We saw that from the very beginning when we said that you can make the most general solution as a product of x and a product of t. Thank you. OK. In fact, everything I've been saying to you for the last five minutes has assume this is independent of x.

OK. So just take five minutes at some point and go repeat what I just went through. And make sure you feel that you have a feel for the fact that this is an envelope only, that at one instant of time, you might see this pressure waveform, you might see 0 all over, you might see this one at any point.

Any questions on this? It's looking at a function of a couple of variables. And the only way to build any insight is just to ask yourself what does this look like in space and time.

OK, by the way, there's something that I think we should have talked about and didn't so far, but it's never too late. You remember that the most general solution-- we actually got to this two ways. One by just taking this solution and plugging it into the wave equation and showing that it satisfied the wave equation. And the other way was to derive it through the exponential and then use Fourier.

But we found out that p of x, t in a plane wave had a most general form of some function of t plus and minus x/c. Then we just take the minus for now, you can do the plus. This is a sinusoidal function. This becomes a function like what you've seen up here, cosine omega t minus x/c.

Now, that function, if you plotted it as a function of x, of course, goes along like this. There's a thing called a wavelength. That's what we're getting to.

You've used it, you've heard about it, I'm sure, in many disciplines. The distance between two corresponding points on adjacent periods of a periodic function. It could be this point here, any point, it could be this point here. And that's generally called lambda.

And how do you get lambda? Well, this is an x. What's the increment in x that you have to go through to make the periodic function? And lambda only has meaning with respect to periodic functions. What's the increment in x you have to go through to make the argument increase by 2 pi? Because that's the minimum x that you can go until the thing repeats.

So what is it? Omega, that minimum x that we have to go through we'll call lambda. Omega lambda over c is equal to 2 pi, or lambda equals c 2 pi over omega, but omega over 2 pi is f. So this is c/f. And the form you're used to seeing all of this in is lambda f equals c.

And don't try to remember it. A good way, by the way, to not remember is to 10 times tell yourself I will remember lambda f equals c. But you should never remember it. You just go back to a periodic function and bang, it falls out immediately just finding out how much you have to increment this to increment the argument by 2 pi.

OK, now, let's look at another thing. We saw a similarity of networks to waves, one-dimensional waves. Or the similarity even could be in more than one dimension if you arrayed your networks in space in two or three dimensions.

There are a lot of similarities. You can look in here and you can ask what the impedance is, just like you look into a network and you ask for the impedance. If you're in the electrical domain, this would be, let's say, well, there's voltage across the terminals and there's current through the terminals. And z would be the complex amplitude of v/i.

We have exactly the same thing here when you look into this tube, right in here, and ask what the impedance is. Where are the two terminals? Here, it's easy to grab them. Where are they there?

Just think of what the meaning of impedance is. If I push my hand against the air, move it fast, there's a pressure generated on it. So I have two variables, I don't have two terminals in this physical structure. But move my hand, that's the velocity, and I get a pressure.

So there's a couple of kinds of impedances. They only differ by an area factor. But the first one is called in acoustics specific acoustic impedance. And it's defined as the ratio of pressure at any particular point, p of x, s-- impedances are always ratios of complex amplitudes, they're not time functions-- over U of x, s.

Let's look at the impedance in this tube. What we're really doing now is taking what we talked about in the first lecture, the Professor [? Chu ?] approach, just taking the simplest thing, one dimension, and looking at all aspects of it until we begin out of all these equations to build an insight. And then the rest is going to be-- I shouldn't say much easier, just easy.

So we're looking at the impedance into that tube. Let's see how we can do it now. I have pressure and velocity, I take the ratio of these two. Is that OK? I don't even have to do it? Well, then I can skip that part.

You can't take ratios uptime functions and never get an impedance. Take the ratio of this function to this and you'll get a very interesting looking impedance. We've got to go back to your basic expressions for complex amplitudes. That's what it's a ratio of.

Now, let's see, I had done this-- darn it-- last time. I think I got the-- yeah. See, what I have to do to get the ratio of the complex amplitudes for this particular example of the tube, is I have to plug in my boundary condition. My boundary condition, it says P-plus is P-minus. So let's do that. And we're dealing in impedance now, we're looking at impedance for our periodic functions for our sine waves at any given frequency.

So let's see. We'll do this for Z. I'll leave it as x just to remind you I'm looking at it in sine waves, I'll put a J omega in here. OK, let's do it. So I'll try to do it in one step. Now, I'll plug this fellow in here and get rid of the P-minuses. So I'll have P-plus of s times e to the J omega x/c minus e to the minus J omega x/c. And downstairs, I will have, again, this P-plus, this becomes P-plus, factor it out, I have a P-plus over Z0. P-plus-- what a mess-- of J omega over Z0, e to the J omega x/c-- be careful now, how did I get this plus, this is a plus sign- minus e to the minus j omega x/c.

OK, the P-pluses go out. The impedance, by the way, if the-- yes?

SPEAKER 2: [INAUDIBLE] on the bottom, you have the e to the j omega x/c, omega negative x/c reversed now.

DR. AMAR G. BOSE: Oh, yeah, yeah, yeah. Thank you. The minus up here, plus up here. Thank you, wow.

Now, when you get this, any expression for impedance. If the complex amplitude of the excitation appears in there, go back and start over. Because impedance is a property totally of this box or of that tube. It is not dependent upon excitation.

And so these, of course, canceled out, just as they do-- you never found an impedance for a network like this. Let's say, suppose in the box there was a network like that, the impedance better not depend on the complex amplitude of the voltage or current.

OK, so let's write this thing down. Z of J omega is equal to J omega and x, or x and J omega. Z of x, J omega is equal to Z0, this comes upstairs-- OK, that's the same thing. I'll just copy it down once more. e to the minus J omega x/c plus e to the plus J omega x/c. And downstairs, e to the minus J omega x/c minus e to the plus J omega x/c.

Now, you look at that thing and you know what's going to happen. This is almost what-- almost a cosine, missing a factor of 2. This is almost a sine, missing a factor of minus 2j. In other words, the sine would be e to the J omega x minus e to the minus J omega x/2J.

So we see that, and let's see if we can do it. We would need a 2 under this fellow to make it a cosine. We'd need a 2J under this fellow and a minus sign. So the 2's would go out, so I have a minus J down here, so I have to put a minus J up here. If we come out of this right the first time, we'll be lucky.

Let's see, minus J up there, cosine, omega x/c sine omega x/c. OK, I could take that and I could say that's equal to JZ0. Put the J, when I have it downstairs, it goes upstairs. And as a plus J, just think of those things as vectors and you'll be always very straight. Minus J is a vector going down this way. And if you think of that, you don't have to remember that minus J goes to plus J.

This thing then is cotan, omega x over c. Now, remember one thing which I tend to forget regularly. All of this is good for x equal to or less than 0. That's the only region for which the whole problem is defined. So x is always negative in this expression.

OK, so let's plot that. This is the Z. Now, when you have something like this to plot, Z, generally, what you do is you plot J times it. Then you're plotting just the real quantity on the graph.

So J times Z of x and J omega is equal to minus Z0, multiply this by J, I get minus Z0, cotan omega x over c. But x is negative, so that's equal to plus Z0 cotan omega magnitude of x/c.

And what's it look like? If I plotted the JZ, well, let me start over here. I'll plot this as a function of x first, x being now negative. And I'll plot here JZ, that's Z.

Cotan omega x, OK. x equals 0, a cotan is this distance over this. So for x equals 0, it's infinite.

Then it starts up here again, goes down here again, starts up here again, and repeats. This is the way that function looks. Let's see. I'd have to move this argument by 90 degrees pi over 2 to get to the first place. So that point here would be omega x magnitude, x0 magnitude I'll just call it, over c is equal to pi over 2.

And so x0 magnitude would be pi over 2, c over omega. c over omega, well, let's do it the first time. We found out from over here that we had lambda f equals c.

This reminds me of Norbert Wiener's two daughters. Weiner had purchased a parrot because he had leading mathematicians and physicists visiting his home over here in Belmont regularly. And the daughters spent an enormous amount of time and taught the parrot, whenever it decided to speak, to say that M equals F-A.

So Wiener came in one day and he heard this thing. And he's spent hours and hours and hours trying to retrain the parrot the other way around and never succeeded. So lambda f equals c.

Let's see. So x0 is equal to pi over 2. Now, I can convert this to omega. Let me do this all in steps the first time. x0 equals pi over 2c, and omega is equal to 2 pi times f, and c.

OK, so the pis go out. This is 1/4 of c/f. And c/f is lambda, 1/4 lambda. So I can put this down as lambda over 4, lambda over 2, et cetera, 3 lambda over 4 on up.

So when x equals the corner of the wavelength to go for the first zero et cetera. Now, I can plot that, and I would like to do that as a function of frequency also, for a given x. Let's say x equals x1. And x1 obviously is a negative number, but it doesn't matter. I have absolute value in here now.

If I plot it as a function of frequency, it goes like this, omega 0. You can see that. And you can figure out the places here that it goes through 0, namely when this argument, again, is pi over 2, that is the omega-- let's do it. I'll call it omega 0 for the moment. Omega 0, and x1 is given over c, is equal to pi over 2. And that's when you go through the first minimum.

So omega 0 is equal to pi over 2, times c over x1. pi over 2, c over x1. And so on, twice that, three times that.

Now, that's how the impedance looks. It never looks that way in a network. In an RLC network, you don't see impedances that go on periodically. So it's a different shape, different contour than you've see. But the concept is the same.

And now comes a really interesting part. I think if any of you were asked at this stage to tell why and when a resistor is a resistor, or why and when the capacitor that you pick off the shelf is a capacitor, and justify that out of Maxwell's equations, you'd have a rough time. Because Maxwell's equations, you've been taught, are for waves, and Kirchoff's laws and some VI relations are for circuits, and the two shall never meet.

Well, they do meet. And we'll see it right now. And what we're seeing is exactly the same for electromagnetics because you have the same transmission line equations.

Let's just expand this time function. We're going to approach this in a couple of ways. But let's expand the cotan function, wherever it is.

So JZ of x, J omega is this fellow here. We expand that, it turns out to be-- well, there's a Z0 in front of it, of course. But cotan, I remember, I think, the first two terms. The first term is the upside down of the argument, c, omega x, if you have in this case an x1, minus 1/3 of the argument right side up, omega x1 over c, plus-- and it goes on down here. Higher powers in all the Taylor series expansion.

Now, it says something very interesting. If omega x over c is small enough, we'll have to see what we mean by that, then this whole impedance behaves like this term. How small does it have to be? Well, if this thing, omega x over c is much, much less than one, this'll be much, much greater than 1, this will be much less than 1 and the higher ordered terms go down.

So what does this mean? Omega over c, what's omega over c now? Remember, we just did it. Lambda over 2 pi, 2 pi over lambda. Don't make me figure it out again.

Give me that thing in terms of lambda. 2 pi over lambda? OK. 2 pi over lambda x1, much, much less than 1. So x1 is much, much less than lambda over 2 pi.

Now, this thing is going to pop up again and again and again in different forms. It's going to pop up in near fields, far fields, when you try to radiate the waves. In other words, x1-- rubbish. We were solving for omega. OK, hold it a second. This thing has to be less than 1, so x1, it's all right. Much, much less than 1/6 of a wavelength.

So for frequencies in this tube, if I move a distance that is much less than 1/6 of a wavelength, we found out that this point here corresponded to 1/4 of a wavelength. So if I'm small compared to 1/6 of a wavelength, I'm somewhere maybe in here looking at the impedance. Small compared to 1/6 of a wavelength.

And frequency, what does that mean? Small compared to 1/6 of a wavelength. If I just looked at the end of the tube-- where's the tube? If I looked in the end of a tube of some given length and I was going to look for frequencies for which this distance was a small part of a wavelength, what kind of frequencies would they be, high frequencies or low frequencies? Very low frequencies.

Because this has got to be small compared to a wavelength. So the wavelength, instead of looking like I drew it up here, it might look like something like this, way down over here. So for low enough frequencies-- and that's what we see up here-- for low enough frequencies, for frequencies much lower than this point, it says, amazingly enough, that I can just look at this term.

So that in that case Z for x distances, much, much less than lambda over 2 pi. Or you can make the equivalent frequency statement just from the same kind of argument. Z is approximately equal to Z0 c, where c is the velocity of the sound, omega and x1.

Now, if I told you, for example, that this was electrical engineering now and you'd never heard of acoustics, and I told you you had an impedance-- which this was J Z0, so I'll just put the J down here instead-- If I told you you had an impedance which was Z0 c over J omega x, 1, in electrical, what would you say? I told you that this is what's in the box when you look in here. What would you tell me was in that box?

SPEAKER 3: A capacitor.

DR. AMAR G. BOSE: A capacitor. All right. The value of the capacitor, unfortunately we have another C now. Capacitor would be, normally it's 1 over J omega c. So the big C is Z0 little c. Velocity, and this is the capacitance. So it would behave exactly like a capacitor of value this.

So what it says is when you're dealing in waves but you're looking at systems which are small compared to 1/6 of a wavelength, you can use lump parameters. And of course, what is this thing? This first term here is a 1 over J omega. And so what it does is it starts way up here also. Only thing is it doesn't come down to 0 when this one went down.

But it turns out, and I think you'll have a problem to look at that, I've exaggerated the difference. If you want 5% error, for example, you have a good region in here that you can use in which this 1 over omega, this capacitor, comes very close to this. So the fundamental consideration, then, in deriving lump parameter theory, and we'll have more to say about this later, is that the systems you're dealing with are small compared to 1/6 of a wavelength.

Now, you look at this and you say aha, I've gone to a single element. Let me make it not look so bad, eh. It gave me a pretty good approximation in here, it sure as heck didn't give me a very good one there. I wonder what would happen if I tried to make a better network approximation to the impedance looking in there.

Now, how might I do that? I might say well, look, this fellow sure as heck has a 0 right here at that frequency. So suppose I decided to synthesize a network which had-- let me see if I can use some-- this is the complex plane now. It has a pole at the origin. That's what the 1 over J omega gives you, rational functions whether-- we know rational function is a ratio of polynomials. And wherever the root is here, it's a pole because the thing goes infinite. Wherever the 0 is, it's a 0.

So if I stuck another 0 where this is, at a frequency given by whatever that is, pi over 2 c over x1, a radian frequency, then I would have something that might look even more. Because at least if we go through 0 here, it might not match exactly in here, but you could nail it down here, and it's going to come closer in this region. That's the driving point impedance of another network.

By the way, I'm doing what you assign me, OK, this part here now. That's a glorified term for what we're going to do. There's a heck of a lot to network synthesis, but we'll talk a little bit about it.

You figured out for me if it was 1 over J omega, there was a capacitor in there. Can anybody figure out what you might stick in that box to give me this pole 0 pattern, which I like because it's a pole at the origin and a 0 here in frequency.

SPEAKER 4: Inductor.

DR. AMAR G. BOSE: Huh?

SPEAKER 4: Inductor.

DR. AMAR G. BOSE: Yeah, where would you like it? I know it's in the box.

SPEAKER 5: Parallel.

DR. AMAR G. BOSE: Parallel with the c. Now, let's see, what does this thing have at the origin? What's its impedance at the origin, at 0?

SPEAKER 6: 0.

DR. AMAR G. BOSE: 0. Ah boy, strike out, eh? This is infinite. All right, let's try it.

Now, there's another way to get a hint at this. What's the impedance of a capacitor at the origin, at 0 frequency?

SPEAKER 7: Infinite.

DR. AMAR G. BOSE: Infinite, 1 over J omega. So it better be in series a capacitor, because anything you put down here now doesn't spoil that. It's infinite at the origin. So it sounds like that's a good bet.

Now, forget about this network now. What's the impedance of an impedance which is represented by these poles and 0's at infinity?

SPEAKER 8: Infinity.

SPEAKER 9: 1.

DR. AMAR G. BOSE: Infinite?

SPEAKER 9: 1.

DR. AMAR G. BOSE: 1? Let's see how you get it. Pole, 0, 0. Now, if we're way up here on the axis-- let's see, we have a rational function now that looks like this. This is s here, or omega, whatever you want. I'll just put them as S's. And this is a 0 at a s minus s1, s minus s2, let's say this one and this one, they're conjugates of each other.

So what happens? you go to evaluate an impedance at any frequency. You can regard them exactly as vectors, from the 0 to the point s. This is a vector from the point s1 to the point s. In this case, they're on the axis, so I have a 0 vector from this one, I'll call that s1, I'll call that s2. I'd have another 0 vector from here up to this point, that s, which in my case is J omega, and I'd have a pole vector downstairs up here.

So as I go up and up the axis, I have two vectors from 0's, this one and this one, that are getting bigger and bigger proportional to my omega. And I have one pole vector downstairs that's also getting bigger. What does the thing go to as we go to infinity? Yeah, it goes to infinity, right? So what's an inductor do as you go to infinity?

SPEAKER 10: [INAUDIBLE]

DR. AMAR G. BOSE: It's a migl. Yeah, migl. It goes to infinity. So this looks like a good bet. Now, it matches the conditions at 0, it matches the conditions at infinity. The impedance is infinite at infinity, infinite at the origin.

Now, this, you should know, has a 0. And it has a 0 when the impedance of this is, obviously, the negative of the impedance of that. In other words, 1 over J omega c is equal to J omega L. Or omega squared equals 1 over Lc is the point with a minus sign, you have two J's up there. So omega is equal to plus and minus J square root of 1 over Lc, which you've seen before.

So when the impedance of the inductor and the impedance of the capacitor are equal, you have a 0. And that's what gives you these two 0's. So looking at this pole pattern now, when it just had a pole at the origin only, you could see it was a capacitor. When it has a pole and then a couple of 0's, you can synthesize a network for it that represents it. About the highest you need to go-- yes?

SPEAKER 11: [INAUDIBLE] be the second term of this Taylor series?

DR. AMAR G. BOSE: Where is that? No. I've got to find it, though. Oh yeah, here it is. No, not really. Because if I just added the second term of that, the Taylor series, I don't wind up with a function which goes through 0 even here. In other words, this doesn't approximate the cotan function. And that thing all the way out doesn't even--

So what I've done, very artificially but very practically, is said, god, I want the lump parameter approximation to this thing. What's the best seat-of-the-pants way of getting it?

Well, I had a pole at the origin. Now I'm going [? to nail to the ?] curve to go through 0 at the origin. And with a little bit of luck, which I can always check, the curve might even be able to be fit a little bit better in here.

Now, there are a lot of networks that you should be able to synthesize by inspection just from the way we have looked at it here. And most of them are two element networks, RLs and RCs. In other words, let me draw for you the kinds of networks that if you give just a few minutes of time to you can begin to see.

A resistance and a capacitance in series. Turns out, that has a pole at the origin, the impedance. Because there's something in series which is infinite impedance. So it's a pole at the origin, has a 0 at s equals minus 1 over RC on the real axis. When this impedance is the minus of this, it's clearly a 0. This impedance, 1 over J omega c, or 1 over sc, is minus R at this point.

This, you should be able to synthesize that. What's that impedance at the origin? Quick!

SPEAKER 12: R.

DR. AMAR G. BOSE: OK. So no poles, no 0's. There's a pole in that network, poles just like you had in RC network, when they go into resonance. There's a pole in that network when this impedance is the minus of this. That's when the currents run around in there. No current out here, but they run around in here and give you a heck of a voltage out here. Exactly when this impedance is the negative of this. So same point, pole.

First of all, a pole of impedance-- remember, Z in electrical is V over I, complex amplitudes. A pole, when this thing's infinite, says with no current at the terminals, you can have voltage. So with no current in here, I can have plenty of voltage across here because once it gets going around, these impedances are the negative of each other. Zip. You get plenty of voltage, just like a parallel LC.

So this network, just take a look, you don't have to remember these things at all. This network at the origin is R because this is 0, L. It has a zero when this impedance clearly is the negative of the other one, and if you just write sL plus R equals 0, you'll find minus R over L for that network.

And you can always check these things. And it's the checking part that, I've said in the beginning, the first day, this constant checking is what's going to condition you to work the majority of the problems that you have in life, which aren't quantitative at all.

You go to infinity. What's this thing when you go way up to infinity? It's the vector from 0. This is R plus sL as you go to infinity. That's the vector from here to the point s. And as you go up, because it's an upstairs vector, that goes to infinity. So does the impedance L which is in series.

So you go to parallel R and L. Now, at the origin, that thing is what, the impedance? 0. The inductance is J omega L. So it's got a 0 there, but where this impedance is the minus of that, namely R over L, it's got a pole.

So as I go to infinity now, the vector from the 0 and the vector from the pole both go to infinity. So the thing goes to finite. And in fact, the finite is R. A pole 0 representation doesn't quite represent the impedance, because a pole 0 representation gives you these factors, 0's upstairs, poles downstairs.

But there's a constant out here. In other words, the constant in this case is such that when you go to infinity, the thing for this one, if I write this one, it's only one 0 and one pole. If I write the expression product over the sum of those impedances, it'll turn out to be one 0, one pole.

Well, as you go to infinity, those two vectors, the one from the 0 and the one from the pole are the same length. And this therefore goes to 1 so you have to pick the constant such that the impedance goes to R. And so the constant in this case is R out from there. You can get it all, the constant, everything, from matching the pole 0 to what you see in the network.

So series RL, parallel RL, and of course series LC and parallel LC. If you have those things that you can do by inspection, it's amazing how much of other things you can do. There are volumes written on network synthesis. But most stuff it isn't really usable. It's of academic interest, but somewhat limited practical.

Because there are ways, for example, to synthesize very complex functions with many 0's and many poles. But you would never want to do this in a real situation for a particular reason. And that is if you had this whole box full of many poles and zeros and you represented it with network elements, an error in any one of the elements can make the whole thing go way off. It's for errors, it's for tolerances of elements that you don't do that.

So what you normally do when you have a complex situation, let's say, you'll break it apart somewhere. And you'll say, OK, I'll synthesize just this part. This is much simpler.

I get myself a network. Maybe it's a driving point impedance, and I have an active device, let's say a transistor, and I put that driving point impedance in here, then I take the voltage off here and make another, let's say, current source and feed the second part. This is number two and this is number one. Number one is here, number two is over here.

And so it's called cascade synthesis. Then you get simpler networks, then the tolerances don't affect you. so it is amazing just from this simple combination how far you can go in practical network synthesis, not by the seat of your pants, but it's just by inspection.

OK, let's go one step more complicated over here. I think you've done it to me. I spent a lot more time than expected. Let's see, let's go one more step over here and say, well, now I've got the yellow approximation which goes through 0 and it's infinite here and it's pretty darn good in here. It's not quite that good. let's do better.

Or suppose I wanted to use my approximation now up to this point and maybe beyond. In other words, when I only represented that tube as a capacitor, or as 1 over J omega in acoustics-- we're going to find out what an acoustic capacitor is later-- I could only use it up to 1/6 of a wavelength and that would correspond to frequencies up in here.

Put the 0 in, I might be able to use it almost to the 0. Put the pole in here, it's infinite there, I might be able to use it somewhat down here. So by putting more elements into the box, you can make the box have the driving point impedance that approaches the real driving point impedance to higher and higher frequencies. So this is about as complicated as you would go by inspection. I'll see if anybody gets it. Don't expect to.

0, we went to the 0's before up there. Now, I'm going to put the pole in. These are all equally spaced. Any idea what you would put in the box? Even partly what you would put in the box? If you can tell me just one of the elements and where I'd put it. Capacitor in series, excellent. Because we've got a pole at the origin and that nails it.

By the way, what does this impedance approach when you go to infinity?

SPEAKER 13: 0.

DR. AMAR G. BOSE: 0. Sure. Three poles, pole vectors are downstairs, 0 vector's upstairs. 0.

OK, we have a capacitor in series. That's for this fellow. What gives you two poles on the axis?

SPEAKER 14: Parallel LC.

DR. AMAR G. BOSE: Parallel LC. This fellow right here gives you two poles on the axis. He gives you a 0 at the origin by himself, but I don't care about that, because I'll wipe that out for this fellow.

So suppose I just stuck that in to see where we go. Maybe it's right, maybe it's wrong. Let's see. What does this network do at 0 frequency? Just checking the endpoints, this network at 0 frequency is a pole, because of this. What does it do at infinite frequency? 0. This fellow goes to infinity, J omega L, this one, 1 over omega c, goes to a short circuit. This one goes to a short circuit. So short circuit. So that checks out.

These poles here, that's fine. These are the poles of this. Because when this goes infinite, the whole impedance goes infinite. So just 1 over root of Lc gives me the frequency of this.

Now the question is where in the heck do these things come from if they do, the 0's? You can do this for the R's and the c's or the L's and the R's, anything. Take a look at just the Lc network.

This fellow is J omega L. Suppose I just plotted for some omega here the vector that was this impedance. It would be having a length omega L. And it's in the s plane. In the s plane, it's a J omega L, so its 90 degrees up.

This fellow is 1 over J omega c. So he is 90 degrees down. So 1 over omega c.

Now, as I raise omega, this one gets smaller, this vector. This one gets bigger. At some point, namely when omega L's equal to 1 over omega c, that's the resonant frequency, these two vectors are equal in length.

But any frequency below or above this unusual frequency where they're both equal, the net is either one like this-- if I go higher in frequency, the impedance of this thing is smaller than the impedance of this one. In other words, higher in frequency, this goes up, this comes down. So parallel, the thing that's dominant, is the one that's small.

OK, in other words, if I have two things in parallel, big resistor and a very small resistor, if this fellow is 100 times this resistance, what's the resistance looking in here? 1% error if I throw this away, right? Why? Because all the current went down here. Put a voltage across these two, all the current goes down the small one.

So when I go to high frequencies, this one just goes way up. And this one comes way down in impedance. So all the current's going down here.

So the minute you go above resonance, at that frequency, whatever it is, if you only needed that frequency, if you were dealing with a sine wave into this network that was just at one frequency, you could replace the whole darn thing by a capacitor-- not this capacitor, but a capacitor, which was the net of the two. If you go below resonance frequency, this one becomes very small. This vector goes down, and this one goes up.

So the capacitor has a high impedance. No current goes down there, current goes all this way. So at any point below resonance, even when there's current in both, the net current, because the resultant vectors-- if I plot admittances here, I get the same thing, the resultant current, that's the best way to do it, actually, instead of impedances-- the net current will be that of an inductor.

So this whole thing below resonance is replaceable by an inductor, above resonance by a capacity at that particular frequency. Well, there is some frequency in which, lo and behold, you have now at that frequency a circuit which is some other L. It will certainly be below the resonance of this. Because only then I can replace it at this frequency by an inductor.

So below this resonance, I will have some frequency at which I have this circuit. And that circuit will have a 0. And so the 0's are supplied by this network. And you can get all the equations you need to get them, because the pole is directly an equation which relates to the poles of this. And then the 0, you have to write another equation for. And then you have a third variable, that's the gain on the thing.

So that's the most complicated network that you would synthesize, let's say. That's not bad, by the way, three poles and two 0's. So this is not the body of network synthesis, but I think a body that will be very useful to you at some point. And again, when you get into acoustic circuits and mechanical circuits like we will, you could do this for those as well. You could make a mechanical circuit that has a given pole 0 constellation.

Basically, this happens very often in engineering. You're working out a function that you would want of frequency. You get it in the form of a rational function. You get the poles and the 0's. And now you say, let's build it. And whether it's mechanical, and you're using masses and dashpots and whatnot, or electrical doesn't make any difference. You just build the network.

Can you ask any questions on this aspect of it? Yes.

SPEAKER 15: So by adding more poles and 0's, we're no longer limiting ourselves to 1/6 of a wavelength. Is that correct?

DR. AMAR G. BOSE: That's right. Because now you've got some sort of an approximation here which nails this point, it nails the point at infinity, and it nails this point. So by golly, you're probably good over a region like that. That's the only reason you'd want to go to this kind of trouble, if you wanted to make a network which simulated the wave situation.

And in fact, maybe I'll do just this much. What else is on here? Let's see. Was I not supposed to do the standing wave? Oh, that was what you have in here. We're not going to do it. We won't make it.

But we'll take another look at that tube to give us some insight. There were two equations that we developed when we developed the wave equation in one dimension. One of them was from Newton's law, partial p with respect to x is equal to minus rho 0, partial of U with respect to t.

Then we did a gas law. And then we did this thing called a continuity equation. And if I remember correctly, the thing we got out of the solution of the continuity equation and the gas law was this, partial of U with respect to x equals minus 1 over gamma-- I hope this is right. Yeah, P0, partial of U with respect to t.

SPEAKER 16: [INAUDIBLE PHRASE].

DR. AMAR G. BOSE: Excuse me. Can't hear you.

SPEAKER 16: The second equation. I think it's partial tau, partial t.

DR. AMAR G. BOSE: Oh, yeah. No, sorry. Partial of p-- this one? Partial of p with respect to t. Yes, thank you.

Looking at these equations-- it's OK?

SPEAKER 16: No, It's partial tau, which is like [? a 1. ?]

DR. AMAR G. BOSE: The partial of--

SPEAKER 16: Tau.

DR. AMAR G. BOSE: When we solved the gas law and the continuity equation, I think this is what we got. Anybody want to check it. This wasn't the continuity equation or the gas law equation, it was the solution of the two of them.

Now, just looking at these two, we can get a very interesting other view of what goes on in a transmission line or a tube or a plane wave. Partial of p with respect to x. Maybe since most of you are EE people, what I'm going to do is I'm just going to make the analogy that these were voltages and currents. If you're mechanical, make believe they're forces and velocity.

If they were voltages and currents and if p happened to be analogous to the voltage, you'd see a thing like this. v, voltage with respect to x is equal to minus rho 0 partial of the current, if this is analogous to voltage. And this, of course, would be analogous to current. I will do this because of two reasons, initially. One is just to show you for maybe the first time or so how you can transfer information from one discipline to another, and the other is so you feel more at home at the moment. Partial of i with respect to x is equal to minus 1 over gamma, P0, partial of v with respect to t.

Now, this is all differential. Suppose I look at a little section dx long. This is the section of the tube that's dx long. Here's the tube. I'm just taking a little section somewhere dx, very, very small.