# Amar G. Bose: 6.312 Lecture 06

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DR. AMAR G. BOSE: On Thursday, we will give out, hopefully, tickets for the LSC movie Stand and Deliver in the evening. And I'll be there to introduce it and to tell you something about the real teacher, who, by the way, is portrayed with absolute excellence in the movie by-- I think it's James Olmos who was nominated for an Academy Award for that film.

The movie itself is free. So if you don't get the ticket, you could still go. But there are reserved seats for this class. And that's what the tickets are for.

One of the most useful things you can do in any subject is each week when you sit down and do the homework, just mentally go through the subject topic-wise from the first thing you did to the second thing to the third thing. And as you'll find out as you do this, somehow it will become clearer, the whole flow.

You look at this subject, for example, we started with gas law, Newton's law, so-called continuity equation, we derived a wave equation, both in one dimension and then in the vector form, which gave it to us in any coordinate system. From the wave equation, we then said all right, we'll specialize to the one dimension. We're going to build our insight in just that. We solved that equation and found out that any function of pressure that was a function of t minus x/c, or plus x/c, depending on which way it was going. It was a general solution.

We got the fundamental building block of any p of t as an e to the s, paren t minus x/c. And from that, we were able to just write down almost the so-called transmission line equations. And from the transmission line equations just comes an enormous body of knowledge. And you'll see insight.

Today, we're going to do something that's based totally on transmission line equations. And it's only algebra. It's simple algebra. But what I want you to see is the process by which you go from what you start with, the transmission line equations, to the end. It is not obvious, and that is the way that new things will be done when you do them one day.

Don't think that you don't have it all together if you can't go from what you have to what you want in a straight line. You really have to be willing to sit there and dream and look and say what if and try things and have them not work and bounce around until all the parts of the puzzle come together. And so really, what I want to show you today, though it's useful, by the way, and it is a way of measuring impedances, that's secondary. The probability that you're ever going to have to measure one of these impedances is very small. But the probability that the thought process will in some way be useful to you is very large.

Let's just look at a tube of the type we've looked at and talk a little bit about this. And maybe it'll motivate why the first person who did this had the idea to do it. We've looked at a closed tube. We found out that the pressure had a maximum at the end of the tube and had an envelope like this.

And by the envelope-- well that expression p of x, t, if you remember, was 2A. A was the amplitude of the cosine coming down on the positive direction. 2A cosine omega x/c cosine omega t plus phi where the incoming wave was A cosine omega t minus x/c plus phi. And this was the envelope dictated by this. It came to 2A at the end.

And the cosine wave, of course, is the wave that you would see at any one point in here if you looked with a microphone at that point. You'd see an amplitude of a cosine wave on the oscilloscope that high. This high over here, this high here.

Now, just suppose you put some sine waves in here and you looked at this thing and then you decided that, gee, look at it with other loads, if you want to call this a load, a short circuit, a closed tube, you might look at other loads. For example, you might do something like this. You might have a tube going on, but you might fill this with some kind of fiberglass, or some generally light substance of a different Z0. This might be called a ZL.

If you did that and you looked at what happened over here, you would see that the peak no longer occurred here. You might get, for the pressure, for example, depending on what the load was, you might get an envelope that looked like this. And it might not go all the way to 0. I might still be going down there, but-- however it goes.

In other words, in the particular case that we studied, closed tube, or it turns out an open tube, a wave comes down here, hits, nothing absorbs. A wave of equal amplitude goes back. When a wave of equal amplitude goes back, there's a point where you always have a 0. No matter what time is, you always will have a 0 because this fell to a 0.

But if the amplitude of the wave going backwards is not equal to the amplitude coming forward, and it won't be more, it will always be less, because-- we haven't gotten there yet, but-- wave carries energy. And the energy is going to be related to the square of the pressure. So the wave's going down this way, hits something, as long as there's no sources in the something, what's going to come back is either equal or less. So it doesn't cancel the wave that was going forward.

So you notice that a couple of things happen. If you were to just play around in the laboratory with different loads, you would notice that the peak changes position, and you would notice that the minimum is not 0. It takes on different values. So suppose, now, somebody gave you the problem of I want an easy way, or a way, at first, to measure the impedance, this what we now call specific acoustic impedance, which is ratio of pressure to velocity. I want to measure that impedance. Well, you might play around in the lab and notice that these peaks were moving in position and the ratio of the maximum height to the minimum height was changing.

You might think to yourself, well, let's see. If I want to measure impedance, in general, that's a complex number. Our Z0 over here in our transmission line turned out to be real. But there's no reason why this has to be real. So to find ZL, there are two-- it's a complex number. At any given frequency at some impedance. So to find the complex number, you need two. You have to find a real part and an imaginary part.

So you just might get the idea that, well, peaks are moving around, ratio of max to min is fluctuating. Maybe there's something in that that I can get a hold of that might make a measurement. And that's really the only motivation that the first person who did this probably had. It wasn't done in acoustics, by the way. I believe it was done in EM, electromagnetics.

So now, what actually happened was you want ZL, but you wind up, in this method, introducing something called a reflection coefficient, which you can go back and forth to ZL. This is the interesting part of how it comes out. But you don't have the reflection coefficient. If you only had that, you'd have ZL.

So now you go over from here to something called a standing wave ratio, which is going to be the ratio of the maximum of the envelope to the minimum in the standing wave. And something called Dmin, which is the distance from where the new Z takes place. So the distance back, in this case, well, if this was the first minimum, it would be this distance, distance Dmin from the place where we choose our origin-- and we know now to choose the origin where the boundary is-- the min will be the distance back to the first minimum.

If my rough drawing here were the case, this would be Dmin. And the ratio of, I can call this, [? P of ?] x, s max magnitude-- we'll get to all of this in a minute-- max over [? P of ?] x, s min will be called the standing wave ratio. We'll introduce these, but just to give you an idea that there are two variables you would have observed that are moving around as you change ZL. Maybe you can measure this and go back to this via something here. And I think you can already see this isn't the way that you would have thought of first.

All right, everything that we can know about waves in a tube, one dimensional waves, comes out of the transmission line equations. They represent the best thing that you can do with a wave going this way and a wave going this way. Exponential wave, sure. Any time function you want, a sum of them.

So let's start with the transmission line equations. p of x, s is P-plus of s, e to the minus sx/c plus P-minus of s. The complex amplitude of the minus going wave, e to the plus sx/c. And U of x, s is equal to-- we remember that U-plus, which I would put U-plus e to the minus sx/c. But U-plus was P-plus over Z0-- where Z0 was row 0, c-- e to the minus sx/c minus P-minus of s over Z0, because this is U-minus, and U-minus is minus P-minus over Z0 e to the plus xs/c.

OK. So the first thing we're going to do is we're going to introduce gamma, called a reflection coefficient. P-minus over P-plus, appropriately named, eh? See P-minus, when you launch a positive wave going here, a negative wave will never a occur. A negative going wave will never occur unless that positive going wave hits something and generates a negative going wave. It's just like you throw a ball against the wall. Nothing's going to come back unless there is a barrier there.

And so you need to have something to generate P-minus. So P-minus is really what is reflected. And that's why you called gamma reflection coefficient.

Now, be careful. Gamma is, in general, since P-plus and P-minus are complex numbers when you specify s, this ratio is complex, so we could let gamma equal-- just to remind ourselves-- some gamma magnitude e to the j eta.

So looking at the problem now, you're in the phase of just stumbling around. Why in the world if you wanted to get ZL do you introduce something else which, like a ZL, has a real and imaginary part, but you don't know it either, doesn't help much? Well, it's going to help in a bit.

So we want to look at what happens at this boundary, because that's where all the interest is. That is where the negative going wave is generated. At the boundary, ZL, which is the impedance here of the second and ongoing part of tube, ZL is equal to Z, which everything I write is for this part of the tube. Because all I'm interested in doing is finding out what the impedance is at that boundary.

So it's just like the other closed tube or open tube in that sense. These equations are not holding for over here at all. They're holding only in this space. So Z of 0, s is the impedance. Let me write it down. P of Z, s over U of 0, s.

Now, I want to just review a second. We found that if you only have a plus going wave, the ratio of the complex amplitude of P-plus to U-Plus , is Z0 in that left part of the tube.

But there's another impedance when you have waves going both ways. And you can look at any point you want in here and look this way. And that impedance that you see there is the ratio of this whole quantity to this whole quantity. In other words, it contains the reflected wave as well as the direct wave.

OK, so let me insert gamma into these equations and then compute that Z. To insert gamma in there, what I'm going to do is simply express P-minus as P-plus times gamma. Let's write it out once. I think I'll put all the steps on the first time. e to the minus sx/c plus gamma times P-plus of s times gamma, e to the plus sx/c. And that takes care of pressure with gamma in there. I eliminated P-minus like that. And this one would become P-plus of s over Z0 e to the minus sx/c minus P-plus of s times gamma, which is P-minus , over Z0 e to the plus xs/c.

OK, now, whenever you're computing an impedance, the first thing you'd look at is make sure that your driving signal didn't enter it, other than the frequency of it, of course, the s. But the complex amplitude of the driving signal has nothing to do with it. So these are going to go out, as you can see.

So ZL would be equal to-- let's see. The P's are going to go out. Z0 is going to come up top. And ZL is defined as 0 for x. It's the boundary between the two media. So the x's go to 0. Exponents, the exponentials become 1, and we have gamma plus 1 over gamma minus 1.

So we can invert this set. By the way, is anybody just-- I'm just curious now-- if you have seen equations of this form and you know what they're called, complex variable? You've seen it but don't remember the name? That's all right.

SPEAKER 1: [UNINTELLIGIBLE]

DR. AMAR G. BOSE: Yeah. It's involved in things like-- yeah. But there's a particular name for that kind of equation. Transformation? Bilinear transformation? Yeah, it's not important. But it's interesting.

You can invert these right away. You don't have to go through the algebra. You can-- ZL over Z0 minus 1 over ZL over Z0 plus 1 is gamma. If you just solve that, invert it. These are complex numbers, so you would expect that to be.

Interesting property, by the way, on the side. Although-- no, no, no, we need it. I told you that we haven't talked about energy yet. But the wave going this way carries energy in it. Clearly, I mean, it can move something out. When you launch a wave going down here, you can move the diaphragm of a microphone with it. So wave carried energy.

And as long as there were no sources to the right at this border, and as long as ZL doesn't equal Z0, there's going to be a reflection. But the reflection, then, always has to come back smaller than the original wave, or equal to it, depending if there's absorption on not in this part. So gamma magnitude, just on that basis, gamma magnitude must be equal to or less than 1. Because the magnitude of P-minus must be equal to or less than that of P-plus.

But you can also see this-- in dealing with complex functions, draw pictures at every point you can. And it's amazing what pops out when you do that. Look at this.

This number happens to be a real number. Do you know from network theory, I hope, that the impedance, if you have a network in a, box and there's some Z here, if there are no sources in here, it's a linear network, that the real part of Z has to be positive? No? Wow, OK.

Think of just making this at some frequency a Z here, which is imaginary, J something and a real. The real part, I could make a resistor at that frequency. If the real part were negative, you could easily compute that if you put a current through here, you'll get power out of it. Not very good. So very nice if it were real, but you don't have those things on the shelf.

So basically, it turns out that if what's in the box is passive, no sources, the real part of the impedance looking in will always be positive. So if you think of that, here's the s plane. I want a bigger s plane. Here's a ZL over Z0. We know it has to be in the right half. We know that this is J omega and sigma. So the ZL has to be here.

Well, ZL plus 1 is this vector, ZL over Z0 plus 1. And ZL over Z0 minus 1 is this vector. So as long as this point is in the right half plane, i.e. as long as ZL has a positive real part, then this vector is always longer than this one. So that means this vector, in magnitude, the length of it is longer than this. So gamma is always less than 1. It's equal to 1 if this point is on the axis, if you have a pure imaginary, as we saw before.

Because remember, when we did the closed tube, P-plus was equal to P-minus . Well, if the point is here, then of course you have ZL plus 1, ZL minus 1, and they're equal in length. So you can see again just from some pictures, you don't have to minimize or maximize quantities. Just look.

OK, so gamma is equal to or less than 1 we've seen from a couple directions. But really, the fundamental direction that was behind all of them was energy. Because it was energy that told us that the real part of this quantity here must be positive. And it was energy that told us that the reflected wave must be smaller in magnitude, or equal to that of the incident wave. Yeah?

SPEAKER 2: First line where you [INAUDIBLE] ZL, at the top there, shouldn't it come to [INAUDIBLE] equals Z0 times the 1 plus gamma over 1 minus gamma? Not--

DR. AMAR G. BOSE: Yes. Thank you. I inverted it correctly, but I'm playing [INAUDIBLE] now, but by accident, getting the right answer by making a big mistake. 1 plus gamma is here, because when I plug in for P-minus-- what the heck is this thing doing? Oh yeah, I already plugged it in. P-minus , I plug in P-plus plus gamma, do the same thing down here, and so this is 1 minus gamma downstairs and 1 plus gamma upstairs. 1 minus gamma. Thank you very much. And this still is the right inversion of it.

OK. Now, great, but we don't have gamma. And so we still don't have ZL.

Now, with some more stumbling, which is what you would do at this point-- you'd probably have introduced gamma, by the way. I'm thinking of how you might have gone through this in the first place. Gamma isn't the first thing you would have done. But eventually, you'd get to the thinking that, well, gee, it's the ratio of the reflected wave to the incident wave that governs everything that goes on back here.

And so that's sort of significant. And that's a parameter which might lead to something. That's about all you do, and that's about how you do it.

Now, let's look at the expression for P-- better yet, the one on the top-- and let's look at maxima and minima for that. So let's take P x, s is equal to P-plus of s times e to the minus sx/c-- top up there-- plus gamma times e to the plus sx/c. That's P of x, s, the complex amplitude of x and s, anywhere to the left of the boundary x equals 0.

Now, remember, I told you that if you had played around experimentally, the envelopes that are drawn up at the top might not go to 0. You would see that they wouldn't go to 0 with some loads. I've got to find the maximum of this and the minimum because those seem to be the quantities that vary as I play around up there. So I want the maximum of this quantity and the minimum of this quantity.

All right, let's see if we can't try something like this. Let's factor out e to the sx, because I'm going to look for the magnitude of this thing, and I want to get everything out that I can get. We'll see in a minute. P-plus of s, e to the minus sx/c. 1 plus gamma e to the plus 2 sx/c.

Now, all of my impedance measurements are always relative to sinusoidal signals. There is the complex amplitudes of these. So I'm going to have different frequencies that my measuring device can generate, my source.

So I'm going to let s equal J omega. Ooh, a mistake. It doesn't go up. Let s equal J omega. And then P of x, J omega is equal to-- and the only reason I'm letting s equal J omega is because I know I'm going to hook a sine wave generator onto the left end of the tube. P-plus of s e to the minus J omega x/c, 1 plus gamma e to the plus J2 omega x/c.

Now, I want the max and the min of the magnitude of this. So let's first write the magnitude of x and omega. That's equal to the magnitude of this quantity here. The magnitude of a product is the product of the magnitudes. So it's the magnitude of this times the magnitude of this.

But the magnitude of this is imaginary. So it's a unit vector spinning around as you move in whatever direction, x or omega. So the magnitude of this is 1. And the magnitude of this product is the product of the magnitudes. So all of this boils down to P-plus of s magnitude, because the magnitude of this fellow is 1 times the magnitude of 1 plus gamma e to the J2 omega x/c.

OK, all we need to do is find now the max and the min. Because the ratio of the maximum of this quantity-- maximum as I move x. In other words, I'm now going back here, let us say, with a pressure microphone. And I'm looking over the tube, I find the maximum, which in the yellow case would be here, and a minimum would be there. I want both of those.

And I want the ratio of them, because the ratio of that max to the min is what we defined right up there as the rho, which is going to be called the pressure standing wave ratio. Pressure standing wave ratio. Or in electromagnetics, the voltage standing wave ratio, or the current standing wave ratio. In any discipline where you have waves, this is the kind of a thing you'd get in one dimension.

OK, I want the max and the min of that. Again, complex numbers, don't maximize or minimize the stuff. Look first. Draw a picture.

Along here, we have-- never mind this fellow first. Because when I look at this, the minimum's going to have the same one. In other words, I write an expression, I look at the maximum of this quantity, then I look for the minimum. This is a constant outside of both. When I take the ratio, this fellows going to disappear.

So what I'm interested in is this fellow right here. So I come out here at 1 unit. Let's make it a little bit bigger. Just a minute. I have to make it-- get some room here.

OK, I come out here to 1. That's this vector, if you wish. This is sigma x. Now I have the vector which is complex, has an angle eta, as we defined it over there, times this. This fellow is a unit vector of angle 2 omega x/c.

So the angle here, now, must be equal to-- I take the gamma e to the J2 omega x/c and that's equal to gamma magnitude e to the J2 omega x/c plus eta. Now remember, that x is equal to or less than 0.

So this angle then is eta plus 2 omega x/c. That's the angle of this quantity. The magnitude of the vector is gamma magnitude because this is magnitude 1.

So now, just a minute. What do you think the maximum magnitude of this quantity is, for all x in the valid region, x negative?

SPEAKER 3: [INAUDIBLE]

DR. AMAR G. BOSE: Huh, what?

SPEAKER 3: 1 plus gamma.

DR. AMAR G. BOSE: 1 plus gamma. What's the minimum?

SPEAKER 3: [INAUDIBLE]

DR. AMAR G. BOSE: 1 minus gamma. Why? Because as you move x, this thing just moves in a circle. And when it's here, it's 1 minus gamma. The vector has rotated around to here. When it's here, it's 1 plus gamma.

So if I then take the ratio of p, the maximum, what I'm really looking for, J omega magnitude max divided by P of x, J omega, magnitude min, which is, by definition, rho, the standing wave ratio-- and by the way, rho, of course, is a real number, the ratio of two magnitudes-- that is 1 plus gamma over 1 minus gamma, as we just saw. The maximum is out here, the minimum is here.

SPEAKER 4: [INAUDIBLE]

DR. AMAR G. BOSE: Pardon me?

SPEAKER 4: [INAUDIBLE]

DR. AMAR G. BOSE: Oh, yes. Sorry. Thank you. The vector length is gamma, that's when spinning around. 1 plus magnitude gamma, 1 minus. Another transformation which you can invert and get gamma is equal to rho minus 1 over rho plus 1.

So if I could measure rho, I have the magnitude of gamma. Now, the magnitude of gamma, unfortunately, isn't quite enough to make it. Because what I'm looking for is ZL. And now I have to get the angle of gamma. Anybody have any ideas how you might, from measurement now, get the angle of gamma? In other words, get eta? What might I--

Take a look at this. Remember I'm going negatively now, if I move x, the vector comes around, comes around, comes around, winds up here at some point. That's the secret. Namely, if I move x, that means the point that I'm measuring the pressure with my microphone, I move back, back, back. Suppose I move it back till I get the first minimum, that's when this vector has come all the way around and points back in this direction now.

And I have a relationship involving eta. And I know what this angle is. If I go this way, it's minus pi. So I'd have minus pi is equal to eta minus-- because x is in the negative direction-- 2 omega over c. And the amount that I had to move x in that negative direction was Dmin. Or Dmin is equal to-- pick all this stuff. Let's see, 2 omega, Dmin-- so Dmin man is equal to eta plus-- go to get these signs right, take this over to the Other side, take this over there-- eta plus pi, then c over two omega, or something. Yeah, looks good.

So I don't want Dmin, I want the eta. That's what I'm hunting for. So eta is equal to 2 omega over c, Dmin-- because that's what I'm going to measure, I can find that-- 2, omega over-- minus pi.

So now, by simply taking the measurement of the distance from the boundary to the point of Dmin, I can find the angle of gamma, which is eta. From taking the standing wave ratio, the ratio of the max amplitude to the min amplitude of the signals going back, I can get magnitude of gamma. I now have magnitude and angle of gamma. And with that, I have ZL.

Circuitous route, but a very practical one. Just some algebra that is addressing a very practical problem, namely how do you take some equations like this and get a simple measurement technique out of it? A technique that's used again and again in different fields. Questions? Yes.

SPEAKER 5: What about positive pi? What if you went to positive pi instead of negative pi?

DR. AMAR G. BOSE: Ah. OK, well, it's pretty hard to go to positive pi, because I can't go to the right in x. x has to be equal to or less than 0. Except I might be able to go all the way around again to some other minimum. That's OK. In other words, when I move x from 0, I've got to go through minus values because all my equations that I built up were based only on to the left here.

Now, what you could ask me is-- I think you are asking me the right question. Positive pi, meaning that why didn't I stop right here? Is that what you're asking? 2 pi? No. I don't know what you said.

SPEAKER 5: First answer was good.

DR. AMAR G. BOSE: OK. But now you raised a nice question, though. Because basically, why don't I go back here and measure this as Dmin. Here I could do that, you see. In other words, I could go to a lesser x and I could come out here with this point. I could write an equation and get it. Yeah?

SPEAKER 6: [INAUDIBLE]

DR. AMAR G. BOSE: Exactly. We'll look at it from the vector point of view too, but let me just tell you the result first. It's not like I really drew it here. What happens is the minimum are always sharper than the maximum in a standing wave pattern.

And you can see this right from here. You wouldn't have to look at it experimentally. If this vector here is near here, and I go D theta in this, I change the resultant vector by a given percentage in length. If this vector is rotated around to a minimum, and just to illustrate, I'll take a a gamma very close to 1, if it's rotated here and I go D theta, I change the resultant vector a much bigger amount.

Let me say it again. When I'm out here, the resultant vector is from here to here. D theta brings it, let's say, this much. There's hardly any change in the length difference here. But if I'm very close in here, if this circle, if the gamma is pretty big, now when I change this D theta, I have a very appreciable change in the resultant vector.

So that's why when you look at standing wave patterns that have minima, that aren't standing wave ratio of 1, you find out that they look like this. This is always sharper. It's easier to find.

Now, when your standing wave ratio gets near to unity so you don't have much ratio between this height and this height and you have things like this, it's hard to tell when it gets really down there. It's always sharper, but when the circle is only this big, or when gamma is very small, i.e. when the standing wave ratio is near unity, then the difference between a delta theta change here and one over here for the resultant isn't much.

And so what actually you have to do experimentally, the way you normally measure this point, is you go to the minimum because that's the better of the two always, and then you go to the minimum and you go back such that the meter reads a given increment each way. And you slosh through the minimum and come to a point that you can read. And then you split the difference between these two points, and you say that's probably the minimum with reasonable accuracy. Yeah?

SPEAKER 7: The expression for the angle there is eta plus 2 omega x/c. And then don't below it says eta minus 2 omega.

DR. AMAR G. BOSE: It's eta minus here. So eta takes this fellow over to the other side which becomes plus--

SPEAKER 7: But see, that right up there, the diagram with the arrow pointing to the angle,

SPEAKER 8: x is minus.

DR. AMAR G. BOSE: Oh yes. In this expression, x is minus. What I do is when I went to the D, I said how much negative do you have to go?

SPEAKER 7: Oh, OK. Oh, I get it. All right.

DR. AMAR G. BOSE: Sorry. That may have confused other people as well. Here, all of these expressions were written for x is equal to or less than 0. Dmin is the distance in the negative x direction. And that's how come it got to the sine here. Yes?

SPEAKER 9: I can see how for each frequency you get now impedance from steady state measurements. But wouldn't it be straightforward to, and maybe it's very difficult, to just look at the pulse response, and knowing the wave propagation velocity, be able to back it out from the transform--

DR. AMAR G. BOSE: In other words, look at a pulse, put the pulse in, and measure what?

SPEAKER 9: Measure the reflected wave that comes back.

DR. AMAR G. BOSE: Yeah, a very narrow pulse, such that it went down and came back. Yeah. Now, the problem with that is if you looked in the time domain, you eventually have to take the inverse transforms of those things and go back to frequency. Because what you want is impedance. So that the time domain pulse, if you put one in, went down, got the receiving pulse, you take the inverse transforms of both. The first one's easy, because it's [INAUDIBLE]. But the second one may not be easy. And from that, you'd have the impedance. From the inverse transform, you'd have the impedance of all frequencies. So it's perfect-- you could do it that way.

Generally, it turns out that you're interested in impedance at, at most, a few frequencies, particularly in acoustics. But there are today, I would say, somewhere in about the '60s or early '70s, in electrical transmission lines, you don't use this slotted line anymore. It was used very heavily prior to that.

But now, you have electronic instruments that will generate these things. It will do inverse transforms. Everything right on the screen, you can read it. So you don't use it. But in acoustics, it still is a useful tool. And very often, you're interested in how over the range of, say, 20 to 20,000 Hertz, how does the impedance vary? And you sample it at a few points. Normally, the impedances of surfaces that you encounter, or materials that you encounter, will change very slowly over a whole frequency range.

The absorbing material that you have either behind these slats or up, where you see the white up top, has absorption, as you'll see later, that is a very smooth curve. Nothing at low frequencies, and when you get above, say, 500 Hertz, it starts to absorb. So it still turns out to be a useful thing, but the other way works. Yeah, any questions? OK.

Let me see if there's anything I want to talk about on this, anything further. Oh, yeah. When you want to measure the impedance of a material, I told you at the beginning, I said, suppose you stuff the rest of that tube all the way to infinity with the material, that's OK. Everything we derive is OK for that.

But suppose I didn't really want to do that. Suppose I had a material that was just this thick, and I wanted to measure its impedance. In other words, I have a material that's this thick that could fit into my tube. And I could call this point 0 and this point x.

Now I have a big problem. What I just derived for you may not work. Anybody know why? Just think about it for a second. Here's the ZL material. Here's the Z0. I'm generating my sinusoids. Yeah?

SPEAKER 10: Would you get a second [INAUDIBLE] for the second property?

DR. AMAR G. BOSE: You bet. Out here, there's another Z0. Let's imagine that the tube continued for the moment or if not, you're doing a plane wave and we had this thing in space. But if the tube didn't continue, it turns out there's some impedance out there, as we'll see later from radiation, it's very small impedance. But there is some, to be sure, let's just call it a Z2.

So what happens now, a wave comes in here, says uh-oh, I've collided with something. A wave goes back. This was a P-plus coming in. A P-minus goes back. Then something goes forward, hits this, says, uh-oh, there's another boundary. So something comes back here. That reflects back here again, and some of it comes out that way.

And so what you measure may have not too much to do with ZL. It's all the reflections that you got now from both boundaries. So in a problem like that, you'd have to take into consideration both boundaries. But you don't know CL out so you don't know what the reflection is to this. You could find out what this impedance is out here, but you don't know ZL yet. That's what you're trying to measure.

So this is a practical case. You have to measure absorption and things like that. What you have to worry about is the following. Normally, impedances you're measuring are [INAUDIBLE] ones because you want to absorb energy. And the way you absorb energy is convert it to heat. That's simple. In acoustics, that's where it goes. When you say you've absorbed energy, you pass a wave through something which generates friction with the particles that are moving, and the friction generates heat.

So now what you do, let's suppose we had a ZL that was this big. And I don't care what I have out here at the moment. Don't know, don't care. Provided one thing is met, that as the wave progresses in here, its amplitude, the P-plus of the second medium-- 2, if you want-- its amplitude as a function of x starts going down.

So let's suppose that the wave-- we haven't talked about absorption yet, but you can physically see this. Let's suppose that it got to this end with 10% of the pressure that it had when it came in. Something gets reflected here. We know it's less than or equal to that magnitude. So that thing comes back with a pressure distribution that looks like this.

And by the time it gets back, it's 1%. In other words, it absorbed 90% going this way. 90% of that for absorbed going this way. So you're now down to 1%. And I don't care what leaks through there at 1% accurate. My measurements aren't that accurate anyway.

So when you're dealing with a material that you don't know and you want to find its impedance, you either put enough of it in there-- assuming it's absorbing-- so that the reflected wave won't come back. Or you find out what's out here and then write the equations for, which you can do, the two borders. And that'll tell you finally what to expect out here.

But usually you don't do the latter. You do the former. So that's a point that you need to worry about.

Now, just to take this over to electromagnetics for a second, you can get an idea. Let's say here's your TV antenna. Down here comes a coaxial cable and goes into the receiver. I mentioned this on the first day, TV receiver.

Now, what happens is if the receiver, let's say this is a coaxial cable, so 75 ohms. This is a character-- say Z0 would be made for 75 ohms. Now, if the receiver, if it's 75, no problem. Because the P-plus is going here, and the equations are identical to the acoustic situation, the P-plus wave sees 75 ohms.

Let's see if I can give you an example of that. If ZL is equal to Z0, if the load impedance that the wave bumps into is equal to Z0, gamma is equal to 0. There's no reflected wave. Gamma is the ratio of P-minus to P-plus. So everything's fine. You get your picture.

If ZL is not equal to Z0-- and it isn't in some of the sets, some of the designs aren't done as well as they should be, and this thing is not equal over the TV band to 75 ohms-- then what happens is you get a P-minus going this way. Well, V-minus if you want to call it that. V-plus came in here. So the V-minus goes back up here.

But now, just think. We have two options here. Looking out of the antenna has an impedance, as we're going to see just next time because we're going to talk about radiation impedance. And it's going to be just like antennas.

If you look in the terminals of an antenna, you see an impedance just like you look in a box. The only thing is the world's contained in the box-- because the antenna and the rest of the world. But it's just like a box.

And so if you look this way and the impedance is matched, if this cable impedance is matched to the antenna impedance, that V-minus wave goes and never comes back. So the only thing that you have is V-plus and you're still in.

But if this antenna is not equal in impedance to the cable, then the V-minus wave that was going back delivers a V-plus-plus wave. That come down here. And now you have something in which V-plus-plus differs from V-plus by amplitude, of course, hopefully, but also by a delay of twice the length of the cable.

And since your TV is going like this, scanning, and this is about 15.7 kilohertz, horizontal scan rate, you can figure out how many microseconds and therefore how much length will give you what sort of a ghost. You see a face in here and then you see another one, right around the outside.

So you need two things wrong to make a ghost from an antenna hook up. You need a mismatch at this end and a mismatch at this end. If your antenna isn't designed well, or if your cable really isn't the impedance that it's labeled to be, some people think, oh, well, I've just got to buy a piece of coax cable, and they'll wind up with a 50 ohm cable and a 75 ohm antenna. Or the antenna impedance varies very much with the structure of the antenna, what impedance it sees, looking out. So two mismatches, you're in trouble.

Now, if you have your TV set here and out the window you have the antenna dropped out, it doesn't make much difference. Because the distance is so small, you'll find that the ghost is negligible. But if the distance big--

Now, another place where you encounter these kind of things is, I'll give another example from electrical. Suppose you were trying to measure the input impedance at high frequencies where you would use cables, input impedance of a piece of equipment you have. So in general, suppose you were trying to measure it with a cable that you connected to it, coax cable, and somewhere over here, you have your generator. And the generator is really measuring its load impedance.

Well, the impedance it sees at this end of the cable isn't the impedance you see at this end unless the cable impedance matched the other. Because otherwise, you get a V-minus going back, which, as we've seen, screws up the impedance that you would see at this end. And just in the hundreds of megahertz, you get into this problem pretty easily.

So you can't do the things you would do at low frequencies where, again, remember on the first day we said wavelength relative to the size of your equipment is everything. And if this length is more than a sixth of a wavelength, or thereabouts, what you see over here is not what you see over there. You make all your measurements and there's no use. But if you're dealing in the broadcast band, where you've got 300 megahertz, where you have 300 meters of wavelength, you can run this cable all over the table, plug it in, and you'll measure at the other end the right impedance.

Any other questions? I have an option-- where I'm going to go now is into radiation impedance. And this is a sort of a subject that scares one when you mention it, because somehow up to this point, you've developed a familiarity with impedance of a box. But now, oh my god, radiation impedance. Well, it's just like a box, as I mentioned before. It's a box in which the world is out there are there are two terminals here. And it has all the characteristics of a regular impedance.

Now, it turns out that the simplest example, and that's really all you need to develop an enormous amount of insight, the simplest example of radiation in acoustics is not in the rectangular case. Because if, for example, you have a tube coming out here ending here, if you radiate out here, it's very complex. The waves spread out in all sorts of ways. So you don't radiate a plane wave from a plane wave going down a tube by any means as we'll see. We actually will have to find out what that impedance is at some point.

But to develop the insight, what you want to do is take the simplest geometry that will do it, and it turns out that that's spherically symmetric geometry. In other words, think that the radiator is a balloon that's just moving in and out, spherically, symmetric, you have one dimension, you can develop a notion. And you can see what many people think is--

I can remember when I first studied, I studied this in ENM. By the way-- I shouldn't say this stuff, you may panic-- I never took a course in acoustics. But you can cross disciplines so easily. When your interest crosses, you can bring your intuition, everything else, into the other field. And then when you play around enough in the other field, maybe your intuition is stronger.

There are some problems today when I look at them, I will think acoustically even if they're an electrical problem. And other problems, I'll think electrically if they're an acoustical problem. It's funny, but your insight is developed by just playing around in that given discipline.

But in this sphere, you have this beautiful case which you can see what happens when you try to radiate. You can see that in some cases, there's a field that moves from this pulsating sphere that has absolutely nothing to do with power radiation. It has nothing to do with what you would hear back there. It has a heck of a lot to do with what I might measure close to this thing.

And that happens in all sources. In the sphere, it's going to turn out, amazingly enough, that that phenomenon of what we call a near field, that has to do with power radiation, doesn't occur in pressure, but it does occur in velocity, in the velocity wave that goes out. In general, it occurs in both. And it dies out very quickly after you go away from the sphere.

That's why I mentioned on the first day that you'll find if you've been reading hi-fi magazines that a good portion of it is pure folklore. Measurements on loudspeakers, for example, will be made in the funniest ways. There's a British chap who reviews loudspeakers who has a 30 foot pole in his backyard and a platform on the pole. And he goes up there and he measures-- well, I don't know how he gets there, he's pretty old at this point.

But basically, he has a microphone up there. And he gets the speaker up there. And he makes the whole measurement. Well, you can measure about anything you want, a speaker like that. And it turns out, as you're going to see, that if you take a loudspeaker and you put the speaker against a wall, or worse yet on a floor, and most homes have that, if you put it there, it changes completely.

What it radiates, the power it radiates out into the room, changes according to the surfaces that are very near it. And all of these hi-fi magazines tell you, well, you go six feet or two meters on axis in an anechoic chamber. Well, this British magazine didn't have the money to build an anechoic chamber, so they figured, well, I'll put it on a pole. That's not reflections. The concept of an anechoic chamber is no reflections. So I'll stick it up there in the middle of the night, no traffic, and we'll make measurements.

Well, whatever you got for a frequency response, as you will see later on, can be off by 18 dB. Well, you would never buy a preempt that was off by 1 dB. And when you bring that thing down from the pole and into a room, it can be off by that much. So the surfaces around a radiator are very, very important.

Now, the radiator we're going to talk about isn't this pole thing, for a moment, it's a sphere in space, space with air, a spherical radiator. But from that ideal case, we'll get insight. And from the insight, we'll be able to see what happens in practical situations, and things like near field homes will absolutely not be [INAUDIBLE].