Amar G. Bose: 6.312 Lecture 19
DR. AMAR G. BOSE: OK, today hopefully we will start and finish what we have to say about microphones. There are all sorts of microphones. Olson, of RCA, the head of RCA Sarnoff laboratories, in his books in the '40s, probably the best coverage of all of the different types of microphones that you can imagine.
But basically, microphones fall into two categories. Microphones they call pressure sensitive. They're ones that whatever the pressure is on the diaphragm, the electrical signal out of the microphone is proportional to that, over the bandwidth of the microphone.
Then there are what people call velocity microphones. Some people call the same microphones pressure-gradient microphones. But from Newton's Law you know that the pressure gradient is velocity, gradient of pressure row 0 dUdt. So pressure-gradient and velocity mics are the same thing. They respond, basically, to the particle velocity. And these microphones are fundamentally different in ways that we will see today.
So let's first just look at the problem of making a measurement. In physics you always have the issue of when you put a probe in to measure something, do you disturb what you're measuring and therefore don't get the right result. So if you had just a diaphragm this is a side view of a circular diaphragm and some electronics here and a signal, a volt. There could be electronics in here and mechanical things in here that we'll see. And you get a signal out.
Well, of first interest is, what does the pressure look like on the diaphragm when you put this into a field? Let's say it's a plane wave. Let's say there was a plane wave of amplitude p1, or let's say, call it p plus, coming to the right, positive x. There is something like p of complex amplitude, p of x and omega is p plus e to the minus j omega x over c. Just a plane wave travelling down.
Well, from things that we have discussed in class, what would you say the pressure on the diaphragm would be at very, very low frequencies? Wavelength is enormous at low frequencies, compared to this diaphragm here. Well, perhaps in a little jiggle in there to remind you that the diaphragm has some compliance.
What do you think the pressure would be on the diaphragm, at very, very low frequencies, if this were the incoming wave? Same pressure. Remember, we talked about sound waves ignoring objects that are small, compared to wavelength. We talked about putting a stick into a pond and then throwing a big stone in and watching the wave go by. It ignores the stick, as long as the dimensions of the diameter of the stick is small compared to the wavelength.
So you would expect the pressure to start off as a function of frequency. Pressure on the, let's say, the diaphragm as p1. Now when the frequency got very, very high, what would you expect? You did a lot-- yeah.
AUDIENCE: [INAUDIBLE] zero?
DR. AMAR G. BOSE: Very, very high pressure coming in here. Wave is hitting here. Somehow I get a feeling that when the wave hits here, there's got to be some pressure on the diaphragm.
AUDIENCE: You'd get a p plus and a p minus.
DR. AMAR G. BOSE: You'd get a p plus and a p minus, because something would hit. You always do get that but at very, very low frequencies. That's so small compared to the wavelength of course, that there's no interference in the incoming field from the reflected field. But now you're getting to the very, very, very small wavelengths. Double, sure. To that very small wavelength, that thing looks infinite in terms of wavelengths.
So you would expect it to come to 2 p1 out here. And now, in between something happens that's similar to what you saw in radiation impedance. This area, in about here, is where the circumference equals lambda, circumference of the diaphragm. And what actually happens in here is something like this- get some wiggles and it goes like that.
If you study the diffraction around the microphone from a wave coming in here, you find some very funny things. It even can go above 2 p1. Just like the radiation of things could go above from a piston, the radiation impedance could go above the actual z0.
Now, I'm going to just take that for a fact. I'm not going to worry about this detail. In fact, maybe I should take a little detour here and tell you, for those of you who are going to become teachers one day, I'll tell you or share with you a little story of the best teacher that I had at MIT, who was professor Guillemin-- network theory. He got his doctorate in mathematics from Germany, under Sommerfeld, who was one of the leading mathematical physicists of the day.
Guillemin was extremely competent in mathematics. And yet he was a teacher who could make things look so simple that it completely-- if he would teach a class and somebody else would teach a class, the students that learned from him just had an insight that was unbelievable. And he had the philosophy that he would some time-- obviously, he was very competent in mathematics, and in his graduate courses, he would be teaching in complex variable theory.
And he would sometimes teach you some theorems in network theory that weren't quite true. He wouldn't tell you that they weren't true. Because the exception was something of probability zero that you'd ever bump into. And his philosophy was that if you get to the point where you actually encounter that exception in something you do, at that point you'll be sophisticated enough to recognize it and know what to do with it.
But if he introduced it, the details of that, then you wouldn't see the-- you'd get so caught up in the mathematics of that that you wouldn't see the whole picture. So he sometimes did those kind of things, that he'd give you a theorem, and he'd prove it, and actually it would be 98% or 99%, in all the cases that you were going to bump into would be true. But there were exceptions.
Well, it turned out-- I was just young on the faculty at the time, and I had taken his courses, graduate courses and undergraduate. And I heard he was going to retire, and that this particular year he was teaching for the last time. And so I wanted to see what the-- I just wanted to go and see this person in action again, one last time. And I went.
It was in building four. And that's a building where they, it was a classroom of about 35, 40 seats. And these horrible seats that all left-handers hate with the board, the blackboard that's the armrest for your right hand and the black table in the front. And Guillemin was there, so I sat in the very back. I had taken that class years before. I sat in the very back, just to see him once more.
And it was a very bright student in there, known to all the faculty at the time, because he was a Greek student by name of Pezaris. And in each of the classes that he was in, he was always at or above faculty level. And he was not a wise guy. Every question or comment that he made was not to show off. It was something that was on his mind, and he was discussing it at the level of the faculty.
Well, this particular day Guillemin was talking about conformal mapping and the general network theorem. And he was going along at the blackboard on the left side of this room. And all of a sudden Pezaris raised his hand and he said, sir, when you have quadrupoles this theorem doesn't hold. Can you address that?
Well, it turned out that was the only time, only situation where this theorem didn't hold at all. And you'd never encounter this in practice. So Guillemin was really, really annoyed. And it was obvious. Because now the whole picture he was painting was blown.
So he turned around, he took a deep breath in, and he started to exhale. And as he exhaled, all of his top teeth came flying out--
DR. AMAR G. BOSE: --followed rapidly by his bottom teeth. And they went boom, boom. And then they bounced from the table onto the floor.
And I was in the back of the room. And I can tell you that there wasn't one single head in that classroom that was visible. Everybody laughed so hard that their head went between their knees. And [INAUDIBLE]. And there was poor Guillemin up there saying, see what you made me do? And literally, when the heads came back up, people, including me I would say, we were holding our ribs because of the pain from laughing so much.
And Guillemin went around the table, and he picked up this teeth. He got out his handkerchief. He polished them off and put them back in and continued as if nothing had happened. He was the best teacher I've ever seen. Not because of that, but in general.
Well, a good lesson for those of you who will become teachers one time is think very clearly about what you want to present. And if there are some details that you don't want to go into, sometimes it's OK. You're not telling the whole story. Or you're telling the slight untruth, if you wish. But sometimes there's nothing wrong with that. And I learned that from Guillemin
And there are things that here I'm telling you that we're just not going to go into this. But there's some things you might not want to go into in class, just to give the person the overall picture. And the other thing that you get out of that lesson that I just related to you was that, for goodness sake, use a good adhesive on your dentures. So in any case, that's the story.
So the pressure looks something like that. Now, let's just look briefly at what kind of a micro--- that's a microphone like this, in which there was only one place that pressure could get to move the diaphragm. It's called a pressure-sensitive microphone. Clearly, what the pressure is on here moves this. And the motion here, somehow whatever means you use to couple, electrostatic or electromagnetic, gets a voltage out of here.
Now, by the way, these things actually do have-- if you look at these microphones, and just so happens one of the people that's doing recordings has a very fancy pair here. This is such a microphone that's quarter inch in diameter. And it's a pressure-sensitive microphone. The electronics are inside the tube. And the electrical signal comes out the other end. They have a little tiny hole in them, on the side.
And the only thing, the hole is there-- remember on the first day, when we talked about sound being a variation in barometric pressure? Well, you don't want the microphone to be a barometer. And so you basically put a little tiny hole in here. It's so small that the time constant for equalization is much lower than any frequency you want to measure. But that prevents it from becoming a barometer. If you take it up to the mountains, for example, you don't want the diaphragm to start moving out of its linear region. And so that's what it looks like.
Well, now the other type, velocity microphone, probably was the first one built. They sometime called ribbon microphones. You can read about them in the text. It's a conductive ribbon here, put in a magnetic field.
And what happens is when you move this in the magnetic field, then you generate a voltage. Well, this ribbon is open to pressure on both sides. And therefore if a wave comes along here, it has to go around the other side, and it creates a pressure on that side as well.
So the-- it's a pressure gradient. In other words, the difference in the pressure from here to here is nothing more than the gradient of the pressure at this point times the distance that you go around to the back side of the microphone. And so these were the first kind of microphones developed. And we'll take a look at both pressure and pressure gradient or equivalently velocity microphones today.
The pressure microphone itself is extremely simple, in terms of its analysis. This is the pressure on the diaphragm. The electronics inside cause the voltage here to be proportional to the pressure there. And we'll see what kind of things that involves. But there's a bunch of mechanical things here you can see.
There's a weight, there's a mass to a diaphragm. There's a compliance to that. There's some compliance of the air inside, just sort of like a loudspeaker. In fact, a loudspeaker is a microphone. If you talk into that thing, you can use its electrical terminals and get a signal out. And you have the model for that, and you can analyze what the signal is. And people have used, in those things, very commonly used in cheap little intercoms. You talk back to the loudspeaker.
Now, let's look at a diaphragm that a plane wave is coming along here. Let's say it's-- well, what should we say? p plus e to the minus j omega x over c.
There's a diaphragm that might be sitting here. And I'll indicate it like this. This is a cross-section of a circle. If I looked head on, I'd see a circle with this border on it. In other words, that's the frame, let's say, that holds the diaphragm.
Schematically, you can look at it this way. This could be, at some angle, theta to the axis, x-axis. This is x here. And this diaphragm has some area, a of the diaphragm. And there's some average path around here I'll call a delta l.
In other words, the average distance that the wave goes to get around to the back of a microphone. Clearly, a wave that's going around to get at the back here goes a little further than the one here. And so the delta l is sort of an average distance.
So then you would say that the pressure-- well, for this kind of a microphone, the pressure on the diaphragm would be-- or if I want to make it force on the diaphragm, I'll say pressure times area. Force in the diaphragm would be simply minus the gradient of the pressure times delta l, times a of the diaphragm to give force.
Why this gradient? Remember the gradient of something-- oh, if there's an angle theta, it's a cosine theta. Oh, no. I'll just take that off and remind you, gradient of a scale there is a vector.
And in the case that we have something at an angle theta, it would be just partial of p with respect to x would be-- I'll do this in a second-- delta l ad cosine theta. Taking care of the sign. In other words, just projected the gradient here on this angle. If the angle we're 90 degrees, clearly you would have no differential pressure between the back and the face, because the wave would be coming like this so that I've taken this into account with the cosine theta.
It's a minus sign here. Why? Because gradient is in the rectangular-- well, think of it in the rectangular first-- gradient is p of x plus dx minus p of x. That's something that, if it were positive, would be a force in the negative direction.
So we're measuring everything with respect to the positive coordinate axis. So it's minus the gradient times the distance. The radiant is the pressure at this point minus the pressure at this point, over the distance between the two. And so to get the difference in pressure, it's gradient times delta l.
OK, now, so much for that. Let's take a look at what happens when a plane wave comes in. This is the wave that's coming in. Just have to look on here, and I get force on the diaphragm is equal to minus the partial of this thing, with respect to x. So that is a minus. And there's a minus that's going to drop out here. So that'll be a plus p times j omega times e to the minus j omega x over c delta l a diaphragm cosine theta. So all of that-- and that's p plus. --all of-- here's p plus e to the minus j omega x over c, which is the incoming pressure.
And if I called x-- let's let x equal 0 at the point of the microphone. So I get rid of that. I get then that this is j omega over in-- what happened? I had omega over c. Yeah. --j omega over c p plus delta l ad cosine theta.
So no problem except for the business of the pressure variation that occurs here. Normally now, you want to get this circumference-- if you want the microphone to extend to high frequencies, you want to move this point up in frequency as much as you can. And that's why, it's the only reason why really that you go to-- well, that's not true. It's not the only reason-- but you go to diameters that are very, very small.
This mic might be good and it depends on the electronic equalization in there. But if you calculate the circumference of that tiny little thing, you'll find out it's good to at least 30 kilohertz. Depending on the electronics, I could stretch it, because you can equalize for that curve that you see there in the electronics. It's inside, if you wish, and make the microphone go a lot higher than that. 50 kilohertz is easy.
Normal microphones that you see are about a half inch diameter. The problem with making them very small is that as you make them smaller and smaller, the force on the diaphragm is of course proportional to the area. And you don't get much force.
And so you very soon get down to the point where you're in the noise level, electronic noise level of the components inside. So you'll get a hiss out of the microphone with no acoustic signal in. So from that point of view, from noise, you'd like to make it very large. But when you make it very large this point moves down in the spectrum.
Now, what I'd like to do is look at this same microphone for a different kind of incident wave, a spherical incident wave. Why? Won't tell you now, but we'll see it.
So let's take the same mic. We'll say now that it's at a distance r from a spherical source. Has an area, same area as an angle to the radius from the source of theta, same as before, area of the diaphragm. So we have the same relationship. Only thing we have to put in is the gradient for the spherical wave.
Well, the spherical wave that's impinging on this thing is of the form p of r and omega is p plus over r e to the minus j omega r over c, as we've seen a number of times already. So let's plug that into this relationship. And I'll just project the cosine. I'll take the gradient along here and then project this area as the cosine at the end. So I can really use exactly this relationship now with this replaced by r, partial of p with respect to r, because that is the gradient in spherically-symmetric coordinates.
OK, so we get force from the diaphragm is equal to partial of p, with respect to r, is minus-- because r occurs twice-- p plus over r squared, e to the minus j omega r over c. Now, a derivative with respect to the second factor, that will be a minus sign also, because I'm going to drop out the j omega over minus j omega over c. So that would be minus j omega over c, p plus over r, e to the minus j omega r over c. Times all of this times delta l ad cosine theta. Delta l ad, area of diaphragm, cosine theta.
OK, now I can simplify this. Look, I have p plus over r minus j omega r over c, which is just p of r and omega. I could get that out of here also.
So I could write fd is p of r and omega-- or whichever way you want to write it-- times-- let's see. I took a minus. I should have taken the minus partial to the p with respect to r. I took partial of p with respect r. It's in here. And it had to minus signs in it, so this comes out a plus sign. fd was minus the partial of p with respect to r. OK, so that then looks like 1 over r, left over from the first term, plus j omega over c, all times delta l ad cosine theta.
OK, now let's look. It's very interesting to look at what the difference between these two is. They are identical, except for this term. This p of r and omega at the diaphragm could be called p plus. That's the incident wave.
Now just this term is different. So let's suppose now that we had a plane wave on this microphone. And we had arranged all the electronics inside the box so that we get a flat frequency response out. Flat frequency response meaning to the plane wave pressure that was coming in here, the transfer function from the pressure to the voltage that's going out would be flat.
So let's assume that we've done this. We've equalized the thing. So here is the omega. We have a flat frequency response. This is plane wave response. This is the pressure on the diaphragm or the pressure of the incoming wave. I'll just call it p plus. The voltage at the output divided by p plus, frequency response.
Now, let's suppose that instead of that plane wave coming into the microphone, we arranged it so there was a spherical source out there. Microphone's sitting same way it was before. Well, since everything is identical but this term, if omega over c is large compared to 1 over r, the two responses have to be the same.
So it would look like so at high frequencies. But now-- I wonder if I should do this in two steps for you. Let me go over a little bit of scratch paper over here and just do the expressions you see first without-- these expressions just are force on the diaphragm. Then electrically inside or mechanically, we correct for those and make them flat frequency response. But let me do it in two steps.
Let's just take a look at this expression versus this. For the plane wave you've got this. It's a function of frequency. This was force on the diaphragm over-- let's see. Just a moment. How can I do that? Yes. --force on the diaphragm over the p plus, let's say.
Now, for the spherical wave, we had exactly the same thing at the high-frequency end, but as omega comes down-- look at the expression up top there-- all of a sudden omega over c gets small compared to 1 over r, and the thing does this. It gets flat. Now, in the design of the microphone, the electronics and the mechanical stuff, we decide we don't want a microphone that looks like this white curve anyway. That's not too interesting.
So we flatten it out, frequency response. We equalize it. But when we do that, as this is the normal practice in microphones, you always make the response flat with respect to a plane wave.
Now, you put the thing in a spherical wave and you get 6 dB per octave. Then let's look at where the 3 dB point is. 3 dB point, the difference between these is where this quantity, 1 over r equals omega over c. Clearly this is imaginary. This is real.
So at this point where omega equals r c, you have the 3 dB point of the microphone. Now let's just see what those numbers turn out to be. If this turns out to be a tenth of a Hertz, who's going to worry about it. So let's take a look. Omega is 2 pi f. So that is f is equal to r c over 2 pi. Yes?
DR. AMAR G. BOSE: Where are we? Omega over c. 1 over r is equal to omega over c. Omega is equal to c over r. Thank you very much. I would have gotten very interesting numbers out of this one. Let's see. c over r, over 2 pi. Is that right? Yeah. OK.
Now, let's see. Let's take r is equal to 5 centimeters. 5 centimeters, the reason I'm taking that number is because I'm going to talk about microphones in a minute. And that's a reasonable number for a mic being held by singer, for example. 0.5 meters for r. c is 345, so f is equal to 345 divided by 2 pi divided by 0.05. 0.05, that's 0.5 times pi. That's about 1,100, 1,100 Hertz.
So at about 1,100 Hertz, at 5 centimeters, believe it or not, the microphone starts going up at 6 dB per octave. So let's see, that was r equals 0.05. r equals 5 centimeters.
So let's say-- what is it-- that r equals 25 centimeters instead of 5 centimeters. Then f would be, that would be 5 times that. So it's a fifth of this thing, maybe approximately 220 Hertz. r equals 50 centimeters. f is approximately equal to 110 Hertz.
So what you see is that the frequency response of this thing changes depending upon how close you hold the microphone. Now, the voice-- and this is the real bottom line for the whole thing-- the voice is a good approximation at lower frequencies to a spherical source. So by history only, turns out the microphones that people use in all the broadcasting studios today are pressure gradient or velocity mics. And they've been used since the early '20s, when they were developed, till now. And it's almost the next century, and they're still used.
And this is precisely why, when you listen to the news and you hear the fellow say, this is the news- five people murdered in Kenmore Square in a bloody battle. Then you think the fellow's been taking some extra dose of hormones. Nobody sounds like that.
Listen to the radio next time. Now don't listen to a radio that has no bass response and listen through a Hi-Fi system. And what you'll notice is the person's voice isn't real at all. And it's much worse on men, the effect, than it would be a woman, because the fundamental frequency of the voice is higher for the woman. So it's much more noticeable in the men.
Now, in fact, to show you, give you an example of what happened, how this can really hang up people-- and they don't-- this thing is called a proximity effect. It's like some mysterious thing that people know about microphones. But they think that's the way it is, the people in broadcasting, for example.
Now, I can remember when we were developing the first prototype sound system for General Motors. It was going to go in a Cadillac. This was in early '80s.
And I was out in the lab one night. I came through on a Saturday night, late in the evening. And a group of engineers-- some of our best. They were really good. Most of them were from here. And they were really struggling at this late hour. And I think the car was supposed to leave on Monday.
And I sort of looked at what was the problem. And they said, you know, it's terrible. They said, when we get this sound system right for music, it's terrible on voice. And when we make it right for voice, it's terrible on music.
And I went over and I looked at the microphone. What they had is a microphone outside the car. And the friends, the different engineers who knew each other's voice well, would get in the car and listen to the person speak. And then they would try to equalize the thing till it sounded very realistic. And then they'd turn on a CD or whatever source and no, it was terrible and vice versa.
And so what happened was they were actually using a velocity mic. Now, the pressure mic didn't have any problems with that at all. The pressure mic responds to the pressure on the diaphragm. That's it. So I asked them, now, go out back into the labs and go get a pressure-sensitive mic. They came out with a pressure-sensitive mic. And immediately, when you got the voice right, the music was right.
So what happens now, you go into stores, and you'll see people-- the audio files, most of them are dying out now. And maybe that's good. --basically, they'll be saying, now, listen to the voice on that. It's more realistic than on that one. Well, you can be sure that if the voice is more realistic on A then B, A is terrible on music. I can't say what B is, but A will be terrible on music.
So people have these notions that are only, they contribute to the folklore, because they don't understand that there is a reason why the voice doesn't sound good when the music sounds good and vice versa. And it's only by tradition. And tradition is so strong in these fields that you can point this out. It's no big-- this is no rocket science here. You can point this out, and it doesn't make a change at all.
People are so used to, well, that's my voice. And if I change to this other thing, let alone take hormones-- something else they think might have happened-- and I'll lose all of my bass of the voice. And so I can't afford to change. Stations won't change.
Since you're going to bump into this in your career, I'll give you another example. One time the CBS recording studios, out in Los Angeles, called me up and said, look, when you come to California next, we would like to give you a demonstration. That's all they said. And so I said, all right. These were recording engineers.
And so it turned out that the next time I could get there was on a Sunday actually, passing through. And they came into the studios on a Sunday. And what they had, each studio was a small room. It was a room, I would say, about from the camera here, at most.
And it looked like this. It had a huge console here with at least 24 channels. And up here it had four giant-- these things were no more than five feet away-- four giant Voice of the Theater loudspeakers. One, two, three, four, somehow. And these are the ones they used behind the screen in the movie theater.
And so they-- I didn't have any clue what they had me out there for. They sat me down at the chair here, and they said, look, we've made the following recording. We'd like you to balance channel one. That was the one over here.
So I reached over. And I couldn't reach the thing. I went like that. I balanced channel one, and I came back and the whole balance was different. Because of course, when your ear was over here, and you're this close in the direct field loudspeakers, you were getting a totally different balance of channel one than you would get when you sat back here.
When the minute I came back and I started to reach again, then they said, stop. That's what we wanted to show you. And then they said, look. And they had taken two of these 901's that we had at the time and hung them right below these, just two. And then they said, look, I want to switch over, and we want to do it again. And with those, the image stays. For reasons that we can-- well, it's not the subject now. But you move over here, and you move back and it'll stay.
And then they told me, they said, you know, to get the definition that we think we need, we blow, on the average, two of these a week. Now these are theater speakers made for hundreds of people. So and you can now know what they're listening to, or what they used to listen to when they had hearing.
And they said, our head is ringing when we go home at night, from this. Because to get the definition we think we need, we have to play at those volumes. And you do that for 8 or 10 hours a day and you've had it. And they said, with these we don't have to do that. We can play them at a lower level, and we notice that it's fine.
Well, then they get down to what they wanted me for. They said, look, we discovered this. We actually bring our speakers and we-- these were their own, the 901's-- and we do the recording and we take them out. And the reason we do this is we asked CBS in New York. We told them what the situation was, and we said we want to replace these by these.
Then they got back an answer. We have been making records for 30 years, using whatever these were. To them they didn't have any idea what they were. And the records sell. And we're not about to take a chance that if you change the monitors that the records might not sell. So you use what you have been using for the past 30 years. That's the end of the story.
So these fellas were bringing these things in in the daytime. And when any of the New York bosses came, the word got around, and they put them under the chair somewhere and hid them. So that is the inertia that you will find in almost every field that you go into. There's an incredible-- their inertia was coming from a business. The CBS people in New York didn't have the slightest idea what was good or what was bad. All is they knew is dollars were flowing in when they were using this. And they weren't about to risk that dollars would stop if they used something else, anything else. It didn't matter.
So I think I've already quoted to you the Maurice Maeterlinck's statement about progress. No? The Belgian Nobel laureate. A lot of the Nobel laureates have said the same thing, by the way. But he was a Nobel laureate in literature and said it so well.
Maurice Maeterlinck was his name. And he said that every crossroad on the highway leading to progress is manned by 1,000 people appointed to guard the past. And that's something that is really so. And you'll experience it. It's not limited to any particular field. But as you go ahead-- and you're likely to do that, because you will have backgrounds to do that-- as you move ahead in whatever fields, be prepared. When you make something that's a real progress, be prepared for the 1,000 men at the crossroad appointed to guard the past.
OK. Now, so we see the so-called proximity effect, and how it can affect the microphones. If you just change to a pressure microphone that whole issue is gone. There is no so-called proximity effect.
Now, there is another reason that one could say people use velocity microphones. And we'll get to that right now. This, to me, is not a justification to use them in any studio. It might be a justification to use them if you're one of these reporters who, after a plane crash, are going to all the relatives in a big crowd and, what do you think. How do you feel, and all this and that. And you need a directional microphone, where the crowd noise is such that you want to point it at the person and hopefully minimize ambient noise.
So this brings us to directional microphones. Now, to be sure this velocity microphone is highly directional. In fact, you would know enough, I think, right now to make a pattern for that microphone.
It would look like this. Namely, clearly, if it's perpendicular to this, you get a big response. If the plane of the microphone is parallel to the wave coming, you get nothing, 90 degrees and then it's symmetrical on the back. Get a polar pattern like that. That's not such a good one if you want to point a microphone to somebody and then point it to yourself. It-- same difference.
So let's look at the following microphone. And this, I think, is the best opportunity to acoustically model something. I think this is one of the nicest examples. And if you really see this example, you'll almost have fun modeling acoustic devices.
Let's take the thing like this. This is the shell of the microphone. If I were looking in here, the whole thing would be circular symmetry. We have a diaphragm across here. I'll indicate that. These are, by the way, in microphones these diagrams are typically metal, very, very, very thin metal with a corrugated punched like a spider pattern on the outside, so that it can move in and out without moving.
OK. And across here there's a screen. And here's a volume, v0, let's say. And there's a pressure wave coming in here. We'll say it's a plane wave, p plus. Well, I'll do it this way. I'll say p1 e to the minus j omega x over c. And I'm going to declare that this point here is x equals 0. And x comes along here. So that the pressure at this point then is when x is equal to 0 is just p1.
OK, we want the response to that-- Now, by response I will take the force on the diaphragm. I'm not interested in what you're going to connect in here, electrostatics or whatnot.
By the way, what is normally used in microphones like that, and most microphones to date-- unless they're electret, which is a permanently polarized microphone developed by Bell Labs some time ago. But those microphones that you see here that need an external amplifier are a metal plate like this with a perforated metal plate behind so that air can move through. But that plate and this plate have a big potential across it.
And then it looks like this- the plate and the perforated plate. And there's a big battery with and big, big, big, big, resistance across here. And so with that, when the sound wave moves the plate, this plate here, then the current, the time constant of this with this capacitor is very, very large. This might be megohm resistance.
And so in the short time of sound waves, you can consider that you don't have much change in current going through here. And so you get a change in the voltage across the capacitor plates. And that change in voltage is amplified. And that's your output. That's the typical capacitor microphone.
OK, now, let's assign to this an ra, it's an acoustical resistance of the screen. You have a compliance in here. You have an impedance of the diaphragm. If we're going to-- I don't know whether we'll use it, but-- yeah, we will. Absolutely.
When I say z acoustical of diaphragm, this means that just the diaphragm itself, the mass and the compliance and if there's any loss in it, it's all some impedance. OK? Because the mass of the diaphragm, the compliance part is right across the mass. This is earth, as far as the system is concerned. And the diaphragm moves relative to this point. And of course, that's the point at which the compliance is attached both ends. So they're all in parallel. That's some impedance z of the diaphragm.
Now, let's assign-- well, first let's write down, since that relates to what we've done before, let's get the pressure at the other end of this. There really are-- the variables here are, well, let's assign vv2, volume velocity at the second business end of this. And there's a pressure 2 out here. There's a pressure 1 here. And there's a vv1, volume velocity through the diaphragm. So I have one port here with vv1 and p. Another port here, vv2 and p.
Now we want to find out what goes on in this. OK, let's first get an expression for the pressure here in terms of the pressure here. That has to do just like it was upstairs there with some overall delta l, some average delta l.
So the pressure 2 I can get from pressure 1 just by saying it's pressure 1 plus the gradient times of the pressure evaluated at the front surface times the delta l. So p2, the pressure is p1, the pressure at the front diaphragm plus the gradient of the p. This is a plane wave, so the gradient of the p is minus j omega over c, taking the partial of this thing with respect to x, times p1 times delta l times cosine theta.
Gradient of p, if I take this-- I took it at x equals 0, so I evaluated the gradient and then I don't have the j omega-- whatever it is-- omega over c times x hanging around, because x was 0. OK, so I got to have pressure 2. Now, gives me a relationship.
I have now to find out how the volume velocities and the pressures at the two ports are related. And this is where the fun comes in in making an acoustical model. And in this step I want you to ask everything you can that doesn't seem to be obvious to you. Because if you get this, I'm very happy.
Now, a two-part set of equations. We can always write them down in the form. Remember, you saw in networks where you had a current 1 and a voltage 1. Depends on how you want to write this, you can sign it any way you want, i2, v2.
And you could always write these equations as v1 equals some impedance. So z11, which we call i1, plus z12 i2. And v2 equals z21 i1 plus z22 i2. And then you can find out what these things are, hopefully, like we did in other models. Or you could write them the inverse of this set of equations, with i's here and v's here. And of course, these would be admittances then. These would be y's.
This thing is called driving-point impedance, by the way, for a very obvious reason. It's the impedance you see looking in terminal one with terminal two open-circuited, that you'll only see the driving-point impedance, so to speak. And similarly, z22 sticks out. It's the impedance, ratio v to i2, that you see looking in terminal two when you set i1 equal to 0. And these are called transfer impedances, because they give rise to a voltage at a different terminal than the excitation term.
OK, now we want to do the same thing for acoustics in this set up. Namely, let's write pressure 1 is equal to something times vv1 plus something times vv2. Pressure 2, pressure on the other side of the diaphragm, is something times vv1 plus something times vv2, just like this. And we want to get these parameters by looking at that acoustical circuit, just like we got in the mechanical or electrical models that we did.
OK, now without looking there first, tell me what this coefficient means. Just in words. If you can tell me what this means in words, you can get it. Yeah?
AUDIENCE: It's when that coefficient is what the pressure would be, p1, if you had no vv2, which would be kind of like a wall there.
DR. AMAR G. BOSE: Yeah. Don't even think of the structure. Just tell me right in terms of these equations.
AUDIENCE: So when vv2 is 0--
DR. AMAR G. BOSE: When vv2 is 0--
AUDIENCE: --then you have some vv1, what is p1.
DR. AMAR G. BOSE: Yeah. The ratio of p1 to vv1, under the condition that vv2 is 0. And this fellow here would be exactly the same way. It's what p1 would be as a result of having a volume velocity 2. Don't look over there. Ratio of p1 to vv2 would be this coefficient if this were 0. That's all we need to know.
So let's see. Let's go for this one now. vv2 we said was 0. Don't worry about how it's 0. It is 0. We said so over there, from the equations.
So that means there's nothing flowing through here. vv2 is zero. So now we have vv1, and we want to get the pressure here.
What is-- now vv1 is the volume velocity right here at the diaphragm. What's the volume velocity inside here? Hmm?
DR. AMAR G. BOSE: Same. Better be. A heck of a lot of matter piling up somewhere if it isn't. So the volume velocity-- just like a current source, you ask, here's an impedance, and you drive in this with a current source. What's the current here? And what's the current here? It's the same, no matter what the impedance is.
So volume velocity says you're moving air through here, and then it better go on the other side. OK, the volume velocity now we know is the same here and the same here. All we have to do is compute the pressure.
So what do you think the pressure is in here, in terms of everything written down on the board? The volume velocity in here is vv1. See, what I want to do is get the pressure in here. And then I can get the pressure drop across here. And the total pressure here would be pressure in here plus the pressure dropped across there.
So I have volume velocity 1 coming into here. There's no volume velocity here. What's the element that looks like with volume velocities a through variable? What's the electrical element? Capacity. Remember?
Remember, you have a things like this, or whatever shape it is. And there's a thing here. And you look in here, and you have pressure and volume velocity. pressure over volume velocity equals 1 over j omega ca, where ca is v0 over row 0 c squared. We derived for a tube, and then we did it for any shape.
So the pressure then here is 1 over j omega ca times the volume velocity. We got the pressure in here. And then we have to add the pressure drop across here.
And pressure drop across here is easy. We know the impedance of the diaphragm. Impedance of the diaphragm times the volume velocity is the pressure drop, just like r times i is equal to v drop, or whatever you want to call it, v drop across the resistor. i is flowing through it. So zad, impedance of the diaphragm, times the current is equal to the pressure drop across the diaphragm. And the pressure in here is equal to vv times 1 over j omega ca, the compliance.
So we have here zad plus 1 over j omega ca for this coefficient. Due to vv1, this is the pressure inside. This is the pressure across the diaphragm. The sum is the pressure of p1. Finished.
Now this coefficient here. This sticks out when you say vv1 is 0. And it's the ratio of p1 to vv2. Well, if vv1 is 0, the only way that can happen is if the pressure out here, if p1 has to be whatever the pressure is in here. Otherwise the diaphragm's going somewhere. And you have a vv1.
So when looking at the equation, saying, I want vv1 to be 0, and then the ratio of p1 to vv2 is this coefficient. OK? So vv1 is 0, which means that the pressure here is equal to the pressure here.
If I have vv2 coming in here, what's the volume velocity coming at the other side of the screen? vv2. Obviously it's volume velocity. It's a current going through the thing. It has to go through, and so it's vv2 on the other side.
And what do I have in here? A capacitor of 1 over j omega ca, acoustic compliance. In fact, I'll just call this acoustic compliance. You know that it's related to volume by v0 over row 0 c squared.
So what do I have in here? 1 over j omega ca. And it was positive, because if I have a volume velocity coming in here, that generates the pressure 1 over j omega ca times the volume velocity. That must be the pressure on the outside so that you have no volume velocity.
OK, let's go for this term now. This one in here. vv2 with vv1 equal to 0 is the ratio of pressure at terminal two to volume velocity, when vv1 is 0. So vv1 is zero. Nothing going out here.
I want to get the ratio of pressure to volume velocity once again. I get the pressure inside, which is 1 over j omega ca, plus the pressure drop across here, which is ra, acoustic resistance times volume velocity 2. So we have here, ra plus 1 over j omega ca. Any questions? Yes.
AUDIENCE: Why isn't the ra taking into account the first equation?
DR. AMAR G. BOSE: And why isn't the ra up in any partic--
AUDIENCE: Yeah, vv2.
DR. AMAR G. BOSE: vv2. This one here. OK. Now, this term was the contribution of pressure with vv1 equal to 0. So no vv1 in here. So the pressure here must have to be equal to the pressure here, otherwise you'd have a vv1. So it's the pressure here generated by putting a volume velocity through here.
Well, this fellow here now has nothing-- the volume velocity on the other side is the same, independent of what the resistance is, because it all goes through there. And so it's just like the same thing saying, the current here is the same as the current here. So that's why the ra didn't enter into it. Because you put vv2 into this, and the vv2 generated the pressure, p1.
So it never occurred in that. But that is exactly the kind of a question that at least half the class needed to be answered. So I appreciate it.
OK, so the next one, you don't even have to look. Only reason you'd look at the physical situation to find the next one is just to make sure everything was all right. Otherwise you'd look right here and go j omega ca, bisymmetry. But what is it? It's pressure 2 generated by volume velocity 1 with volume velocity 2 equal to 0. Volume velocity 2 equal to 0. Volume velocity 2 equal to 0, and you want to get pressure 2 in terms of vv1. Pressure 2 in terms of vv1 with volume velocity 2 equal to 0. Volume velocity 2 is equal to 0. Pressure here is equal to here. Otherwise there would be volume velocity.
So if you have volume velocity 1 coming in here, what's it doing? It's this diaphragm, just like the argument we went through with the resistor on the other side. Diaphragm doesn't care at all, because same volume velocity's going in here. The volume velocity in here is squeezing this mass. And that's giving rise to the pressure.
And the pressure inside and outside must be the same, because you have no volume velocity here. So that's-- you would go through that argument, in a sense, to make sure you went through the right argument over here and got the same answer. Otherwise what you know is that they have to be the same.
So we have now here-- let's see. p1 is known. We have three equations. p2 is unknown. vv1 is unknown. vv2 is unknown. Three equations we can solve.
And if we do that, we get a-- I copied it down somewhere-- the solution of those three simultaneous equations comes out to be-- oh, if you want to have it in terms of, it depends on what you want. If you want the volume velocity of the diaphragm, all you got to do is solve for vv1. In these you have three equations and three unknowns. If you want to have the pressure across the diaphragm as the thing that you want to look at, because now the pressure across the diaphragm is what's driving the whole differential pressure-- in other words, the net pressure, p here minus here, is what drives the whole thing.
So it depends on what you want to have to look at. But if you want to do pressure across the diaphragm, then I would need a one more expression, which says that's zad times vv1. That's the differential pressure across the diaphragm, differential pressure, net pressure. Net pressure.
So you solve all that stuff, and you get the following. Just solve them simultaneously. You get p1 zad, impedance of the diaphragm, ra plus delta l over c, velocity of sound, times ca cosine theta, and za ra plus ra plus zad all over j omega ca. A mess in terms of frequency, in terms of-- only impedances are there, but I only want this for one reason.
The way the microphone is normally designed is so that-- remember, we said there's a reason for using a pressure-gradient microphone-- it's designed so-- and this is the first structure that you will see this true in this subject. It has a directional pattern. And it has it with elements that are very small. You saw it with big-spaced elements, quarter wave and whatnot.
And it has a directional pattern that's pretty good frequency-wise. Namely, all is you have to do when theta is 180 degrees, when the microphone's pointing the opposite way from the wave, you just arrange it.
This is minus 1. So if you arrange this coefficient here to be equal to this- in other words, if ra-- by the way, it's normally written. I'll take it over there-- so ra ca is equal to delta l over c. Then for 180 degrees this response is 0. And what you wind up with is-- if you just take out these coefficients here, with this condition, you get a 1 plus cosine.
And so you get a directional pattern that looks like this. Theta is 0, meaning that the microphone and the wave are facing each other. Theta is 180 degrees. You get a 0, a cardioid pattern.
Now, the very interesting thing is if you go one step further and you make a electrical model out of this. We didn't do that. We had just the acoustic equations, and we arrived at this.
Now, just go one step further and make an electrical model, and you have a very interesting interpretation of this. This is easy now. Electrical model, let's make pressure analogous to voltage. Got v, just v. And then of course, that means that the volume velocity is analogous to cur. Now we should be able to make that network really very simply.
Total impedance, vv1. When you open circuit this, you look into an impedance, which is zad plus a capacitor. And the capacitor is the mutual term, because it occurs-- the mutual term, you could look at these as loop equations, the off-diagonal terms border between loops. The diagonal terms are the sum of the impedances around the loops.
So you can see right away that what this is is a network zad, capacitor of value ca. This is voltage. This is current. Well, said in another way, these are pressure and this is volume velocity. Think of them either way you wish.
And this fellow here, when you look into the port two, you see this impedance. And that's the self-impedance around the loop. And this is the mutual. So I already have the mutual, which I better have. And this fellow is just a resistor of equal to value ra. So that's-- if there are any questions about this, please ask them now. Because something very nice is going to come out of this, and I don't want you to think that this is a mystery.
OK, so this is the network. This is port two. This is p1, voltage 1, or vv1 here, i1, i2, v2, which are equal to vv2 and p2.
Now, let's look at the condition. It seems sort of funny. You think about how in the heck can you really get a zero when the wave is coming in this direction? Because wave comes in here. It goes through here. It pushes the diaphragm. And then the wave comes around here, pushes the diaphragm.
Well, let's take a look just at the electrical thing. Don't think about any acoustics at the moment. We're looking for a situation in which there's going to be a zero current through here. That's the volume velocity of that microphone. And when the volume velocity of that thing goes to zero, there's no output. Then the volume velocity of the diaphragm obviously goes to zero.
So somehow or another, to get volume velocity zero, that means current zero in this. So the voltage across here, somehow, has got to be equal to the voltage across here in this network. Well, when that happens, if it does happen, there's no current in here. So I have a very simple network, namely ca and ra.
Now, let's look at the transfer function of this network. What symbols can I not-- oh, let's call this vs and v5. OK?
So v5 over vs. It's a voltage divider as you-- oh, 5 and then s-- that's bad-- 6, 6ns. So the voltage divider there is 1 over j omega c, 1 over sc, if you want. ra plus 1 over j omega ca, which becomes, if I multiply it by-- to get it in some standard form, I'll multiply by j omega and divide by ra. And then you get it as j omega plus 1 over ra ca. Multiply it by j omega 1 over ca upstairs.
OK, now look at this. This thing is a pole in the s plane, a pole at 1 over-- so I just let j omega equal s, and that's where the denominator's a 0, where the impedance is a pole. It's 1 over ra ca.
Now, at very low frequencies, frequencies such that omega is small compared to this-- this is a minus axis out here-- you're down here like this. This is the frequency omega. For omega, very small compared to 1 over ca.
This thing in that region, this is about equal to the length here. But there's an angle here. So if you looked at the magnitude and phase, what you'd find out for a simple rc circuit is magnitude goes on like this and eventually goes down as 1 over s. And the phase angle, this is a pole, so it's increasing. And it's a pole downstairs, so the phase of the total transfer function is going down like this.
And you look at the phase here, and what is it? It's 10 minus 1. Theta is 10 minus 1, angle whose tangent is omega divided by 1 over rc, which is omega rc and for omega rc see much, much less than 1. The tangent of a quantity much, much less than 1 is the quantity. So theta is approximately equal to omega rc.
Now, for those of you have had-- and most of you have-- the signals and systems course, you know that when you have a linear phase like this, and you're down in this region here in frequency, you have a linear phase that gives you a delay. And the time delay that it gives you-- you know you can pop this right out from Fourier transforms-- the time delay is minus d theta d omega. And theta is equal to this. And theta is equal to minus this, I should say, because theta of this function here is minus the angle of the pole. So minus d theta the omega is equal to rc.
So if you look at this network now, its first effect at low frequencies is only-- an rc network-- is only to delay the signal. Doesn't have an effect on the magnitude down here. It's as close to 1 as you want to get. But it gives an angle given by this. And so it just delays the signal.
So what would happen if you now coupled this with what you know over there? What's going on here is exactly what goes on when the wave comes in here. It sees the resistance and the capacitance.
And the effect of that is to produce, at this point in pressure, p2 delayed. Same magnitude, but the pressure in here is nothing more than p2 delayed. Just like the voltages on the capacitor, it is the same as the v6 is, whatever it was, v source, equal in magnitude but delayed.
And so you would get zero if the delay that you got out of that, the omega, the rc delay was exactly equal to the delta l delay. That's when you would get zero. So if we had then-- where can I put this? Here maybe. --if we had the delay here, which we found out to be this, rc. It's ca, by the way. I keep dropping the darn a in there, and ra. Everything is a.
So we have ra ca is equal to-- now, what's the time it takes for the wave to get to the other side? It's the delta l, partition of that, divided by c. Or exactly this.
So electrically you can look at this thing and see, hey, that's really all that's happening. Now, if you had knew nothing about electrical and you somehow were modeling this, the notion of an rc circuit just creating when you're at low frequencies only delay and no attenuation wouldn't be coming natural to you. And you might not think of this.
But that's when you combine the two disciplines, you get a very good insight that what goes on in here is nothing more than pressure, which is equal to that in magnitude and delayed. And when the delay there equals the delay around here, the two pressures are equal. The diaphragm goes nowhere. You have a zero. So again, the incredible benefit of modeling. Thanks.