Amar G. Bose: 6.312 Lecture 04

DR. AMAR BOSE: I've been asked by the teaching assistants to make a very what I believe is a reasonable request, and that is that you write neatly in your homework. Because they're trying to spend time not just seeing if you have the right answer. If you're sloppy about your writing, that's not too bad. If you are only going to get graded or checked off on the answer, but they're trying hard to follow the steps. And as some of you will know when you become teaching assistants one day, that is a miserable job if people aren't writing clearly. So please make a big effort to do that. And incidentally, I'll guarantee you that'll be useful later.

Today, we're going to talk about solutions in the one dimensional wave case. And we're actually going to build the framework today that will be the framework that our insight will develop. Up until now, it looks like a bunch of equations, and you wonder, where in the heck is the insight out of all this? What we've done today is basically derive the wave equation and show a general solution to it, a solution which I like to call an exponential building block. The exponential is the key, as you know, to solving of the homogeneous differential equation, because all of its derivatives have the same form.

And the exponential is key to the solution of all linear systems for a very basic reason. If you can find the solution to an exponential, you can find the solution to any time function, because any real time function can be made up as a sum of exponentials. A finite sum in some cases, an infinite sum, which is the Fourier Transform in the general case. So just knowing the solution to an exponential, it's a basic building blocks solution. That's all you have to find, and you can find the solution to any time function.

So we did this for the wave equation, for the one dimensional wave equation, and we got a solution that looked like this. p of x and t was of the form p plus-- oops, P of s e to the s t plus or minus x over c. That meant that you could have a plus or a minus, and what was the difference? Well, we looked into that. If you have a minus sign here, this represents a wave, in this case an exponential with some complex number out here. Don't worry that these are complex time functions, because you only add that to its conjugate, and you have a real one.

So a wave like this with the minus sign was a wave going in the positive direction. Now don't under any circumstances try to remember that. Just look at the argument, how do you follow a fixed point, a given point on this wave form? Well, if x increases-- sorry, if t increases, you have to move x in the positive direction to keep the argument constant. So you have to go in a positive direction to follow any given point in the argument, or therefore in the function.

If it's a plus sign in here and t increases, you have to go the other way to keep this argument constant. So that's how you find out which would be is what. Don't try to remember that a minus sign means plus going and a plus sign means minus going. That doesn't work, except for very short term.

So I'll put this in, and I'll leave it out sometime also, but you'll know what I mean by that. It could be a plus or minus sign. And we-- well, let me just use a minus for now for a plus going wave. We can also write this as p of x and s e to the st, where p of x and s then by definition, by just looking at this, p of x and s is p of s times e to the minus sx over c. I'm just equating these two here, you find that.

So in general, we saw last time that when you put an exponential of frequency s into a linear system-- well, into our wave equation, which happens to be linear-- you get out a solution which is e to the st also times a complex number, when you specify what the frequency is and the position in space. In network theory, you specify the frequency, and there was no element of space. Specify the frequency, you found this complex number, and that's all there is to it.

When you think of a linear network with an input here x of t, linear, output y of t, If you think of this x of t as an exponential building block x of s, in general x of s, it could be a constant e to the st, and you think of y of t as y of s e to the s t, then it must be it's a linear system. It comes out with some complex number times the same exponent that you put in, which we saw from solving the differential equations.

If we think of that, all we need to do is calculate this knowing this. That's your frequency transfer function H of s, all right? H of s is y of s over x of s. Once you know that, you know the output. So the beauty of all of this is that chasing these complex amplitudes around is pure algebra. By going to this exponential building block, you have changed your problem from one that is solved by calculus to one that is solved by algebra.

To go from time function to time function without this approach involves H of t, which happens to be-- these are transformed pairs-- the impulse response, and it involves the convolution integral, from x of t to y of t, and that is not a friendly function.

So you're in the frequency domain when you're calculating these complex numbers that go out in front of e to the st. If you're dealing with waves, you'll have a space function also like this. If you're dealing with networks and networks had been formulated, instead of node one, node two, nodes three as one, two, three in some space variable, you would have seen exactly the same thing, and there would be nothing to get scared about when you go to waves.

OK, just to make sure, we'll repeat something that I believe you've done the equivalent of in your homework, the sinusoidal case. Let's do it for this fellow first. Suppose x of t was, let's say, A cosine omega t plus some phase angle that you're going to put into the linear system, a being some real number. It's a nice real time function. You can express that in terms of your exponential building blocks in the form, let's say, x of t equals x of s e to the st, where s equals j omega. In this case, I want to reduce this to this, or I want to make this equivalent to that. And x of s is equal to ae to the j phi, where a is this a, a real a, and e to the j phi.

Because if you plug this in here, you get a-- in this case, if I want this form to be reduced to this, one way I can do is take the real part. Another way I could do it is make this up of two exponentials for which this applied, and one being the conjugate of the other. And so I get something plus its conjugate is always real. And in fact, it's twice the real part of either. So if I add this function to it with these conditions, to its conjugate and take 1/2, I would also get this. Or if I take real part of this expression, I get it. And real part, as you should have I think seen in network theory, it's not hard to prove. Real part and a linear operation, the real part operator and the linear operator, you can commute them. In other words, if I take the real part of the input, that gives me the real part of the output. That's nice. That's for linear systems.

By the way, when you don't have superposition, when you don't have linear systems, the frequency domain concepts are, let's say, all but useless. More accurately, useless, because you cannot any longer say aha, if I know the response to an exponential, I know the response to any input. Because yes, I can take the input and I can make it up of a whole bunch of exponentials if you give me a function like that or something that I have to make, a time function like that. I can make it up of a whole bunch of exponentials.

I possibly-- except for amplitude vary-- you can't even find the response to one exponential, because even to different amplitudes in a nonlinear system, the response will be different. But even if that weren't a problem, you couldn't add the responses. In other words, I make this x of t up of a whole bunch of different exponentials, but I no longer can say, well, I get the response to each one, and I add the responses, and I get the response to some of the input. So all that goes away. So the whole concept of complex amplitudes in nonlinear systems is essentially useless.

There are a few exceptions which solve a trivial set of almost contrived problems. But other than that, you don't have to worry about it.

OK, let's see, where was I? Real part of ae to the j phi, I think I was doing, e to the j omega t. And let's just make sure that all that comes to what we said, namely this time function. So I can combine this. I got real part. I can take a out, it's already real. e to the j omega t plus phi. And real part of each of the j something is cosine something, a cosine omega t plus phi. All review, but useful.

So by simply specializing x of s to a to the j phi and s to j omega, I can represent a real function here if I use this representation. And the same thing exactly with waves, the only difference is the complex amplitude now is a complex function of possibly both x and s. In this particular example, the complex amplitude was not a function even of frequency. It was just a real constant in a phase shift phi.

OK, this is all review and everything else, but if there's something that you don't feel comfortable about here, either ask now, or if you can frame the question better later, ask in the recitations, ask in your tutorials. But just make sure you have it. It is amazing how many graduates of universities today don't understand this basic level. About two or three years ago, the vice president of our company, vice president of engineering, who was a student at MIT, came to me and said, over this whole year, I have interviewed 250 candidates. And I wanted to hire a very large number, and I could only take 5, much less than we had openings for. Fundamentals like this aren't understood.

So they are basic. Nail them. Make it part of you. Not from memory, don't memorize things. But just review the elements of it. It is totally vital to the solution of system problems in all sorts of disciplines and linear disciplines, which is most of the world as we model it.

OK, now sinusoidal example we put on here. This is x of t. If you wanted to do it for waves, I think you have already done it. But well, maybe we should do it once. We don't need that. So let's say p of x and t is equal to a cosine omega t minus x over c-- one of the solutions that we had-- plus phi. And you want to represent that in terms of your exponentials.

So we could write p of x of t. What we want is p of se to the minus sx over c e to the st. We'll use these things interchangeably. This is the same as st minus x over c in parenthesis.

So what do I have to do with p of s to do that? Well, p of s would be equal to again ae to the j phi. And let's just see that we can work backwards. This would be equal to a and s equal j omega. So p of s would be ae to the j phi e to the minus j omega x over c e to the j omega t. So we put all the exponentials here together, we get ae to the j omega t minus x over c plus phi.

And the real part of that, if I want p of x of t to be this, then I represent p of x of t as the real part of this function. And that's real part here. Real part here, well again, this is real. So if I take real part of this, real part of e to the j something is cosine of the something. So it's a cosine omega t minus x over c plus phi. Then we have it. Just again to review, make sure that we understand that.

OK, now we want to go on to the basic problem of plane waves moving through space. We found the general solution, exponential building block solution for p. How do you get u from that? Well, you can say u, variable velocity, that variable we can go back to the wave equation. We can derive a solution for it. But you really don't have to do that, because you already have a relation, at least one, between pressure and velocity. Remember the first one from the mass, Newton's Law? It was in one dimension partial of p with respect to x is equal to minus rho 0 partial of u with respect to t.

So if we have pressure, we should be able to derive velocity. So let's do that in terms of our complex representation. Let p of x and t equal to p, the complex amplitude, p of s, and I'll take a positive going wave so I just don't have to keep putting the plus and minus sign all over. p of x and s e to the minus sx x over c e to the st would be the general building blocks solution for p in a positive going wave.

Plug that in, partial of p with respect to x is only this part is a function of x. This is a function of frequency only, and that's a function of s and time. So differentiating this with respect to x just drops out a minus s over c. So the left side then becomes minus s over c p of s e to the minus sx over c e to the st. And the right hand side is minus rho 0-- in other words, this went to this, this is going to this-- minus rho 0 partial of u with respect to t.

Now what am I going to do about u? I don't have it up here and that's what I want. What do I know about u? I know it's the solution to a linear differential equation, and as such, it's form must be of this form. I'm going to jump one step here, and then I'll go back and explain is. It's form must be of the form u of s e to the minus sx over c e to the st.

On the side, how do we know that? Well, we know that u of x and t must have the form p-- u of s and x-- x and s, I don't know which way I normally write these things, but it doesn't matter-- x and s e to the st. Why? because the differential equation is a linear equation. e to the st building block solution gives an output if you want. e to st into the solution gives a complex number when you specify frequency and space times e to the st.

Now how do I know it's this particular thing? Well, in any of my equations which derive this, which we did before, just let t go to a new variable, t minus x over c. So if the input to a system is changed to the-- if you let t go to t minus sx over c, or t minus x over c, then the output goes to t minus x over c. So I already know that the form of this is going to be exactly here, u of x and s is going to look like that.

Now if you don't even know that, if that's not clear, what you can do is you can say, well, I know this much, because it's a linear equation. There's going to be something out front which is a coefficient, complex number in general, which is a function of space xyz and frequency, and e to the st. That we know for all the linear equations. Just take of this function and plug it in for here, and it will come out in the end exactly what I get if I make this assumption that it was already the u of x and s was of that behavior. And it's worth trying that. It's worth actually going to your homework and just sitting down for a second and just plug in this for the right hand side here.

So I'm going very slowly over these things the first time, because again, they're so fundamental. OK, so I chose not to plug this form into here. I chose to just give you the argument that I did before that the u of x and a is already in this form, and I can know that by substitution of the variable t to t minus x over c.

All right, so that's the u. I know it's going to be in that form, now let's just plug it in here, partial of u with respect to t. Well, the only thing that depends on t is the last term. So it's u of s e to the minus sx over c e to the st and an s when I differentiated with respect to t. So let's see what happens here. s's go out. e to the minus sx over c goes out. The e to the st's go out. And what do we have? We have that u of s is equal to p of s divided by rho 0 c, two minus signs go out.

So we were able to get u of s without going back to wave equations directly from the expression for p because we have the result of Newton's Law right here. Now what does this mean? You know what the complex amplitude of a voltage means at a terminal. It gives you the amplitude and the phase in the sinusoidal case. Well, here's a pressure wave going along here. Believe it or not, the air sees a pressure if you move it. In other words, you don't feel it, but if I take a plate here and I move it against the air, the weight of the plate in all the things I've experienced dominates. But if it didn't, if I pushed fast enough, moved the velocity, I would find the pressure resisting. And that is the complex amplitudes of the pressure and the velocity, it turns out are related by this. And in fact, this is called p of s over rho 0 c-- whoops, just one. One, one, one one. Let me write it this way. p of s over u of s, p of s over u of s is equal to rho 0 c. And this is defined, it's called the characteristic impedance, z0. Characteristic impedance of the medium it's called.

Now be careful here, because I specifically took a wave going in the positive x direction. How many of you, by the way, have had-- I don't know what the number is now, but the field theory course on waves in electrical engineering? How many? Very small. You remember an expression that looks like that, with voltage here and current here? It came from exactly the same kind of analysis. The two disciplines you can, in many cases, trade equations. And this is called the characteristic impedance. For example, if your TV wire, if you have a twin lead coming down from the antenna, the characteristic impedance of that thing is 300 ohms. If you have a coaxial cable, one wire in a coax shield, the characteristic impedance in 75 ohms of that that you can use in television. And this is what the animal is.

Now however, be very careful, because if you were to do this using a negative going wave, this fellow here would have been a positive. Instead of negative, he would have been positive. Then when I differentiate over here, I find out that the end of it, because of that sign change, that I get the ratio of p of s to u of s for the negative going wave, for a wave going that way, is minus z0. So when you think of the thing as being the characteristic impedance of the media, be careful.

And the reason for the minus sign is that the positive coordinate system is to the right. And so if you're thinking of a wave going this way, I push this way and I get a pressure. If you're thinking of a wave going the other way, I'm pushing this way, but that's minus u to get the pressure. And so normally I will distinguish throughout this subject when I do something like this and use a positive going wave, I will in the future try to distinguish this by a plus sign under there. And that simply means the positive going wave. If it's a much negative going wave, we'll distinguish it by a negative sign. So this would really be a p plus and a u plus for a positive going wave. The ratio of the complex amplitude of pressure to velocity is z0, which happens to be rho 0, the density of the medium without the presence of the sound wave times the velocity of the sound.

Let's see how many boards I have left. Ah, good. Now we're going to derive some equations which are almost going to be trivial to derive, and yet are so fundamental. Out of the equations that we're going to derive will come the understanding of why there's lumped parameter theory and why there's wave theory, and when one works on a problem, and when one doesn't work on a problem.

We're going to now consider waves one dimensional situation, just the x-axis going this way and going this way. So let's write an expression p of x and t is equal to the sum of p plus of s, the one going positive, e to the minus sx over c e to the st, the basic building block for the positive going wave, plus p minus of s, it's another function of this, wave going in the negative direction. So that's a plus sx over c e to the st. And similarly, u of x and t is equal to u plus of s e to the minus sx over c, the positive going wave, e to the st plus u minus of s e to the plus sx over c e to the st.

That's all it can happen in terms of our basic building blocks exponential. That's all that can happen in one dimension for any one frequency, that's all it can happen. You can of course have many, many frequencies in there, and many, many terms if you want.

All of this is multiplied e to the st as we have always seen for the solution of the linear differential equations. So let's begin to get that out. We'll factor out e to the st for everything. p of x and t is equal to p plus of general, I'll just-- sometime as we go on through this, I'm going to begin to drop the s. It's just extra baggage. You'll know what p of p plus means, but we'll keep it for now. e to the minus sx over c plus-- I'm going to factor out this, the e to the st-- plus p minus of s e to the plus sx over c all times e to the st. And u of x and t is equal to u plus of s e to the minus sx over c plus u minus of s e to the plus sx over c, all times again e to the st.

Now, these are the complex amplitudes that multiply your basic exponent. This one here, that is really p of x and s multiplying this. If I specify frequency and position, it's a complex number. That's all. And this fellow here is u of v and s. So as we mentioned in linear theory, the whole name of the game in the frequency domain, which is the easy domain to work in, is getting complex amplitudes. Once you get the complex amplitude, you just multiply it by e to the st. And if the input were the real part of that, you take the real part of the product here and you've got the time function.

If it's more complicated, if there's a triangular wave that's launching this pressure wave, OK, fine. You use the Fourier Transform, but you still calculated this.

So, these are going to be the most important of all. Now, we already know something. We know from up here that for a positive going wave, there's a relationship between p plus and the u plus, and it's z0. I told you that if you plug it in for the negative going wave, and I want you to do that, you'll get a relationship between these that's minus z0. p minus of s over u minus of s is equal to minus z0. So let's go one step further and make use of that.

Then we would have p of x and t is equal to p plus of s e to the minus sx over c plus p minus of s e to the plus sx over c, all of this times e to the st. And u of x and t is equal to p plus over z0 p to the minus sx over c plus minus p minus of s over z0 e to the plus sx over c e to the st.

Now, this represents the most general solution in terms of the exponential excitation that you can have in one dimension. And the complex amplitudes associated with that are inside of here. So you can, as I said, represent the whole thing by p of x and s and the other by u of x and s. If we do that-- whoop, board doesn't go up-- we have p of x and s is equal to p plus of s e to the minus sx over c plus p minus of s e to the plus sx over c. And finally, u of s and x is equal to p plus of s over z0 just from the equation above, e to the minus sx over c minus p minus of s over z0 e to the plus sx over c. These equations here are known as the transmission line type equations.

The few of you that have had field theory will recognize these. You've probably seen them with voltage here. For p you'd seen it of e, and for u, you've seen an i. If you change them to that, you get all the equations for the transmission line from your TV antenna, whether it's twin lead of coax or whatever it is, you get the equations that govern that transition line. And you find out as we'll go-- I'm taking you a little bit ahead now. For example, if you have a TV antenna on the roof and you have 100 feet of cable or something coming down to the basement somewhere, and your television set doesn't look like 300 ohms in there, what happens is a wave comes down from the station, doesn't like the termination, and that generates a reflective wave.

Now if you're antenna isn't also quite 300 ohms, that the reflected wave goes back and says, uh-oh, I can't meet the boundary conditions without another reflected wave. And what happens is the wave comes back down again, and that's why you see a ghost. In other words, you see the image, and then you see the image slightly displaced. That isn't the only reason you see a ghost. I mean, if the transmission from the station to your antenna comes like this, and then over here there's another building that it reflects from, and there's a time delay, you see a ghost due to that. But you can see ghosts also due to improper termination of transmission line. And so if you understand one discipline, it's just amazing how much you can transfer it to other disciplines. And we hope to try to do this in the subject.

OK, I'm going to ask you if you can frame a question at this point. Yes?

AUDIENCE: [UNINTELLIGIBLE].

DR. AMAR BOSE: When we did this, we got rho z0, the ratio p plus to u plus. The ratio I gave rho 0c, I told you that if you started with a minus sign, which meant the-- sorry, if you started with a plus sign, which would be the negative going wave, then what would turn out is that the ratio of p minus of s to u minus of s would be minus rho 0c. And that's the one I'm leaving you to do. And because of that, that's where this minus sign came in. In other words, I wanted to get u minus from p minus, and the ratio was minus rho 0c.

I'm going to ask you a question, but I'm going to ask you to, before you raise your hands to close your eyes. Not yet. The reason I want you to do that is because sometimes people are embarrassed to give the right answer, because they're afraid of what others will think. So I want you to, when I ask you this question, I want you to close your hand and raise your-- lips are flapping without my head in gear. Basically, I want you to close your eyes and raise your hand if you think it's appropriate.

How many of you think that this is over your head at the moment? Close your eyes and raise your hand. Keep them up. OK. How many of you think that this pace is OK? Keep your eyes closed. OK, how many of you think that this pace is too slow? Keep your eyes closed. OK. We got a distribution sort of like this. Too slow, OK, and too fast. So I don't know what to do about it, but these people that say it's too slow, bear with us, we'll go faster. The ones that think it may be too fast at this time, it's understandable.

Try to formulate a question. I know how you're feeling if you're in that mode. I felt this way. I think I may have told you, I got drafted from the E department to do a thesis for a Professor Wiener. And for the first nine months, I never went to see him, because the only thing-- I had all his written material, and if I went to see him, the only thing I could say is, I don't understand anything. And I wasn't going to go on that basis. So I know how you feel, but please ask all the questions you can in recitation and tutorial. Because what we have done, believe it or not, is simple. And if you can just get the right keys that open the door, you'll see it.

With this, we can find out anything we want to know. If somebody launches from that wall, let us say, they launch a plane wave going across here. And maybe it hits something over at this end, and I want to know what the pressure is at this point in space as a function of time, or what the velocity is. We can find it all out, and we're going to go about doing that.

I think what I will do at this time is not worry about these great plane waves to go with an infinite wall, et cetera, et cetera, and tell you that you essentially get the same thing if you launch a wave down a tube. In other words, you have a piston here, and you make this piston move, and that launches a wave down here to whatever is out here. It's the same thing as a plane wave while in this region. And that assumption makes some assumption about no friction at this surface, no viscous friction. And air does have viscosity to it.

And so if you were to plot the velocity in the middle here, it would be larger than it would be maybe down at the end. And if you look under a microscope, this thing looks like mountains. And so it stops the particle motion, this wall on the inside. So we're going to assume that there's no viscous friction at these walls. We'll talk about that later when it's necessary. And so far as we're concerned, at this moment, plane waves are traveling down this tube. And we're launching it in such a way that we're not worried about any waves that might go radially in the tube, or anything else. We'll be able to address all of that stuff, I think, very easily once we understand the simplest case, the plane wave.

So when I drew a tube like this from now on, and we're talking one dimensional, we're just assuming it's the same thing as if you had an infinite wall here going an no tube. So let's take the case that we have a given excitation here, and we'll take me case of a closed tube at the other end. We're wiggling this thing around by some time function, and we want to find the pressure and the velocity at any point inside of here.

That is an identical problem to the problem, and I think we'll be able to illustrate that later, of a transmission line. Depends on which analogy you make, and you're free to make any one you want. Pressure analogous to voltage, in which case velocity will be analogous to current. Or you could make it the other way, pressure analogous to current and velocity would be analogous to voltage.

It's analogous depending on which you make to either a shorted pair of wires that go along parallel, or a coaxial cable which goes along with a short on the end, or an open on the end, depending on which analogy you make, and you put a voltage across the line here. And you want to find the voltage and current on any point in here, same problem, same equations. So let's do it.

Sorry, a closed tube. I'm going to launch a wave, a plane wave down here, that will be p plus of x and t is equal to-- well, first I'll do it just with our exponential building block. And then if we want to make a sine wave out of it, so what? We'll put two of them together, or take the real part of the input, and that'll make a sine wave. So we'll say that this is p of s e to the minus sx over c to the st, in general first for the complex.

This is from my complex input. Later I'll say, oh well, now suppose we take the real part of that, we get a sinusoid or a cosine, and I want to find out exactly what the pressure and velocity is at each point in this tube. Well, I have the relations that give me the complex amplitudes everywhere. All I have to do now is find out-- I even have p plus, because that's the wave I'm going to launch down there, p plus here. And I have to find out what p minus is. And once I have p minus, I've got the complex amplitudes of pressure and velocity at every point in the tube.

So let's see, first I have to pick an origin for the coordinate system. I'm will suggest an origin to you, and if you're like everyone else, myself included, you will suggest another one at some point in your studies or your career. And you will only suggest that other one once. And you'll see why when we go through this. I'm going to suggest that the origin be where the boundary is. This is x, and this is 0. And shortly we'll see why, but even seeing why won't do the trick until you forget and don't pick it there. I've had it happen.

So I need to find what's going on here. Well, let's see. At x equals 0, there's a boundary. Now what do you think that boundary imposes upon the general solution, which is given by the transmission line equations? At x equals zero, what do you think that boundary does? Particle velocity 0, because think of a little particle here, it can't go anywhere when it gets here. so it's 0. So it says that this boundary implies u of 0 and s is equal-- u plus of 0 and s, sorry. u of 0 and s is equal to 0.

So if total velocity is 0, at x equals 0, and here you go. That's very nice, the exponents become 0 and e to the 0 is 0. So what do you have? You have that that boundary condition implies that p plus must be equal to p minus. Now just think for a moment if you chose the boundary which would have been normal first choice at the other end. Then when you came to this end where the boundary was, you have all these exponentials floating around in here, and you'll have a gay time before the smoke finally clears, and you get the same answer.

So this thing implies that p plus is equal to p minus, because these are unity, the exponents are unity. This is 0, and we have a plus quantity here and a minus quantity here, they better be equal. So that boundary condition gave us just that, and now we have for the transmission line equations, we have p of x and s is equal to p plus. I can factor out p plus now, because we already concluded that p plus was equal to p minus.

Oh boy, now I've had it. Jeez. OK, p plus of s, I'll continue to put it for a little while, e to the minus sx over c plus e to the plus sx over c for the complex amplitude p of x and s right out of the transmission line equations, and u of x and s is equal to p plus over z0 e to the minus sx over c minus e to the plus sx over c from the transmission line equations.

So now I have the complex amplitudes p and u anywhere inside at any frequency in terms of p plus of s, which is my given excitation. So if I want the time functions at any point in time, simple. I multiply this by either st and I have the answer. For this, use the st input. This p of x and s times u to the st is p of x and t. Done.

Now let's do it for a sine wave, for an excitation that was a sine wave coming down the tube. Questions to here? See if you can-- yes?

AUDIENCE: [INAUDIBLE] transmission line wave lengths then, if I only have a positive traveling wave, the particle velocity and pressure are intakes?

DR. AMAR BOSE: Yes. If you have a positive traveling wave in air, a plane wave, the pressure and the velocity are in phase. Interestingly enough, when you get to things that are not plane waves, and the most complicated we'll go to is, let's say, a spherically symmetric wave, you'll find out that when you're close to the source, or when the whole thing is small dimensions compared to the wavelength, the pressure and the velocity aren't in phase. And in fact, if you were to take a person speaking, or let's say a loudspeaker, what you'll find out is that there's a lot of motion of air in the neighborhood of the radiator.

And it just goes out and back, out and back, out of phase with the pressure. And it turns out, just like in electrical engineering where the power is related to the product of voltage and current, well, so it is in acoustics, pressure and velocity. So these particles that go moving in and out from me speaking here or from a loudspeaker, don't contribute at all to radiated power, because they're out of phase with each other.

And this is one of the millions of tricks that are used in the audio world, of people making measurements on a product that's very close to the loudspeaker. And if you horse around with the microphone position, you can get about any curve you want. But it has nothing to do with what's radiated from the loudspeaker, or what you hear. So the plane wave case, we're lucky. The pressure and the velocity are absolutely in phase at all frequencies.

Let's see, I forgot what I was up to doing here. Oh yeah, I want to give an example of a sine wave. I think I'll do that over here. OK, so we just have to specialize this excitation to a sine wave. To do that, I'll just write that p plus-- I'll start off with a t. Again, this is review. Normally when you once get the hang of this, you're only going to ask yourself, what's the complex amplitude of the input? What's the complex amplitude of the output? Multiply it by u to the st, take the real part if the real part is appropriate, and you have the answer.

But just to make sure, we'll say that p of x and t is a cosine omega t plus phi. a is some real number, amplitude of the cosine omega t plus phi. So I can represent that as real part of some p plus of s and x and s e to the st. And I have to represent s equal j omega for sure to do that, otherwise I'm not going to get the cosine.

Oh, oh, oh, this is not a traveling wave. This is a voltage waveform. t minus x over c positive wave mega t minus x over c. Cosine omega quantity t minus x over c plus phi, that's about as complicated as we'll get. I'll represent that. s equals j omega obviously, and now what do I do about all this stuff? Well, I'll write this as p plus of s. I know what its x dependence is, plane wave, e to the minus sx over c e to the st. That's this much.

Now what do I have to do such that this expression for s equal j omega is equal to that, well I would say that let p plus of s equal ae to the j phi. If you let p plus of s equal ae to the j phi, put it in here, you get that. Anybody want me to go through that again? No? OK.

Now, knowing p plus of s, we can go right to these equations to get these. So p of x and s is equal to p plus of s e to the minus xs over c, OK? p plus of s is this. So it's ae to the j phi. And then this is e to the minus j omega x over c, right from the transmission line equation. And plus p-- I want to go to the transmission line equations, but with the boundary conditions stuck in, so it's this set over here. In other words, where p plus is equal to p minus, because I had a closed tube. So I've got to watch over that end of the board.

Plus e to the j omega, that should be the first equation. Now I could write these things, u of x and j omega instead of s, because I've already let s equal to j omega, is equal to the same thing, except that it's over z0. So ae to the j phi over rho z0, and now e to the minus j omega x over c minus e to the j omega x over c, and that.

So now I have both p and u. And now to illustrate it for a sine wave, we just have to go the extra steps. What's p of x and t? It's this thing here times e to the st. p of x and t is equal to p of x and s e to the st, for s equal j omega, et cetera. And so all of this is then a e to the j phi e to the minus j omega x over c plus e to the j omega x over c times e to the j omega t. Now this looks like a big mess. There's all sorts of j's here. What do I do with this now? How do I simplify it?

There's all sorts of ways to try. When you multiply e to the power of something, you add the powers, of course, the exponents. So I could take this in, multiply it by this and multiply it by that. And that's something that you will probably also do at some point once. If you do that, look what happens.

Always when you get an expression like this, calm down and think what you want to do, and don't try to do it by memory. Don't try to say, the phi's go into here or something like that, with the omegas, that's all nonsense. You look at this and you say, oh my God, I almost see something. Don't let it go. What do I almost see? A cosine, that's a last thing in the world I'm going to let go. So this thing is 2 times the cosine of omega x over c. It's even real.

And if I put this in here, then that's left over, it's OK. If I multiply this in here, I'm finished, I lost the cosine altogether, because I have a plus and a plus over here. And a cosine is e to the j plus something plus e to the minus j. So this I can write as a times 2 cosine omega x over c times e to the j. And now I'll combine these that are left, omega t plus phi.

So now, I'll specialize now to the sine wave. Suppose that the input were the real part of this thing for s equals j omega. That's a cosine wave. Here it is, up here. So now, if that's true, then I simply say that p of xt, in this case, this was general, but p of x and t anywhere is real part of this thing here, p of x and s e to the st for s equals j omega, et cetera. And that's equal to-- all of this stuff is real already, and the real part of e to the j something is cosine something. So I get 2a cosine omega x over c cosine omega t plus phi equal to the pressure at any point x and t.