TheFourierTransformAndItsApplications-Lecture30

Instructor (Brad Osgood):And letís see, review so Ė oh, Iím on. Man, give me a chance here, will you? Where is the review session? The usual place? Review session today Ė what time?

Student:[Inaudible].

Instructor (Brad Osgood):4:15 Ė and where is it?

Student:Skilling 191.

Instructor (Brad Osgood):Skilling 191. Okay. All right. Let me call your attention to the various announcements on the board, some of which you have seen before. So the final exam is a week from yesterday, next Thursday [inaudible] 11:30 in the morning, Dinkelspiel Auditorium. Be there or be dead. Now, I know there are a couple people who have conflicts, so Iím working out some times with them. Once again, itís open notes, open book. Weíll provide the blue books. Weíll provide the formula sheets. You provide the knowledge, and the answer, and the correct answers. Please make sure your answers are correct. It makes the exams much easier to grade. Letís see. Any questions on that? Watch the website for postings of various things. Iím not sure if the solutions for last yearís final are posted yet or not. Actually Ė you donít have them? Anyway, if theyíre Ė I posted the final from last year. Iím also gonna post the solutions if it hasnít been done yet, so thatíll be up there, and thereíll be other things, too. I think I owe you some homework solutions and things like that that will be up, and other sorts of announcements. So watch the webpage. Watch your e-mail for other exciting announcements. There is a review session today Ė the usual review session, I guess, at 4:15 in Skilling 191, and also the TAs will have the regular office hours next week. Iím gonna have to check on my office hours because I know I have some meetings and things like that that I have to go to, so I have to let you know on that. So again, watch the webpage for other sorts of announcements.

Also, you should have gotten an announcement from the registrarís office that the online teaching evaluation form is open, has been open, and will be open through next week, I guess, so I urge you please to go to whatever the appropriate site is on Axess. Donít leave this room. Where do you think youíre going? Oh, too late. Go to the Ė sit down. All right. Go to the appropriate website on Axess and please fill out your teaching evaluation. Now, every time I teach a course at the end of the quarter, as a public service to the students in the class, I offer a few tips on how to fill out the teaching evaluations because I know youíre very busy, and sometimes stressed, and sometimes are stuck for something to say, so let me give you some sample adjectives you might wanna use, like for instance, brilliant, dazzling, never have I had a richer more fulfilling intellectual experience, and of course, close-up please, handsome. All right. Okay. Dolly back. So now I wanna finish things up with a treatment of inverting the Radon transform, so thatís how I wanna finish things up. And again, Iím sorry that we donít have a little bit more time to do a little bit more detail and go into the finer points of this because itís really Ė and show you actually some of the ways that itís implemented because itís interesting, but there are plenty of ways of following this up, certainly in courses on medical imaging and so on with any Ė and as you know, Iím sure we have a very active medical imaging group in the department.

So this is tomography and inverting the Radon transform. So Iíll remind you of the setup, what the Radon transform is, and why this is an interesting problem, and why the question is formulated this way. So the setup is we have this two-dimensional slice full of gunk, 2-D region Ė letís just put it delicately Ė of variable density, density mu of X1 X2. And the idea is that if you can find mu, if you can describe Ė you donít know mu, but itís variable and the question is if you can reconstruct mu then you can tell whatís inside. So you pass X-rays through the region, and you measure the integral of mu along various lines. So youíre restricting mu to different lines, and thatís actually what gets measured. You think of that as the intensity of the X-ray, or related to the intensity of the X-ray as it exits the region. So it enters with a certain intensity that you know. It gets scattered, it gets attenuated by all the junk thatís in there, and thatís what you measure. And from that, what youíre measuring actually is the integral of L along mu. So you restrict mu to a line, you integrate it, and thatís what youíre measuring. So this is along various lines, along line L through the region. So you measure that.

And the question is if we know this, these values, along all lines, many lines, whatever Ė along lines through the region, can we get mu? Can you reconstruct mu by knowing those integrals? Now you think about this as a transform question, and already thatís not so obvious. Thatís not an obvious step. As a matter of fact, none of this is obvious. It took I think a lot of insight, intuition, luck, whatever to formulate the problem this way and actually figure out the solution because as I say, when you see how this is solved, itís just gobsmackingly amazing. So you think of this as a transform problem. And what I mean by that is with mu fixed so itís unknown, we have a correspondence of lines to numbers given by this formula. Thatís a line, and the integral gives you a number. And that correspondence you think of as a transform of mu evaluated on a line L. That is to say Ė and I write it R because itís the Radon transform Ė the Radon transform of mu evaluated on a line L is exactly that formula, the integral along L of mu, so itís called the Radon transform.

It was introduced Ė I actually knew the history of this a little bit more thoroughly, and I cannot recall it now. It was certainly not introduced in the context of X-ray tomography or anything else. It was introduced for purely mathematical reasons, for interesting geometric reasons. The idea was to sort of study the geometry of a region by knowing integrals of sections through it just as a purely mathematical question. I donít think there were any practical implications that were anticipated or attempted certainly at the time it was introduced. So our question is Ė so you know all these values. All right? You know all these values, and the question is can you invert the transform? Can you find mu, given that you know all the values of its transform? Knowing all values, R mu of L Ė can we invert Ė or another way of putting it is can we invert R? Okay. Now, this looks pretty abstract, so to make this tractable I have to introduce coordinates. All right? I wanna write things in coordinates. And what I mean by that is not coordinates on the plane, not something replacing the X1 and X2 coordinates, but I want a coordinate description of the line, and actually more precisely what I want is I want a way of coordinatizing all possible lines. I want a coordinate description of the family of all lines. I wanna write this Ė I wanna write R mu L in coordinates, so script R.

What I mean by that is I want coordinates that describe family of the lines L in the plane. Now thatís actually not outlandish at all. And in fact, youíve actually seen possible coordinates when you first learned about writing equations of lines, so let me give you an example, not the example that weíre ultimately gonna use, but an example that youíre very familiar with, and thatís what we call at least in America the slope intercept form for the equation of a line, so e.g. for example, you learned to write Ė You can write equation of a line as Y equals MX plus B. Thatís one of the first things you learned. And you know that M is the slope and B is the intercept Ė B and slope M. Now you can think of the pair M and B as giving the coordinates of a line. If I specify M and I specify B, that determines the line. So as the pairs M and B vary, I describe lines in the plane. So MB gives coordinates for describing a line. It describes a line, gives the line. Let me just say describes the line, the unique line Y equals MX plus B. So thatís a set of parameters. Thatís a set of coordinates that describe the lines in the plane. Itís not the best one to use certainly for our problem, and thereís a problem with this set of coordinates because it omits Ė it doesnít give you a description of the vertical lines of infinite slope. So itís not the best Ė not a good set of coordinates because itíll miss the vertical lines. But certainly, as M and B vary, M can vary between zero and infinity, can equal to zero but not equal to infinity Ė zero is horizontal lines. Infinity is vertical lines, so thatís a problem. And B can vary between minus infinity and infinity. As M and B vary over that range, youíre describing all the possible lines through the origin, which should be all the possible lines in the plane, except the vertical lines. So this omits Ė so here again, minus infinity less than B less than infinity, and M bigger than or equal to zero less than infinity describes all but the vertical lines.

So thatís not so good. You can sort of fool around with it a little bit and try to make it better, but for our problem thereís a better set of coordinates. It shouldnít surprise you that some set of coordinates to a particular problem than another set of coordinates. Polar coordinates is better suited to describing problems when thereís circular symmetry because the equations are simpler in polar coordinates when thereís circular symmetry than they might be in Cartesian coordinates. So you wanna choose the coordinates that make the equations as simple as possible, or somehow that reveal the essential structure or symmetry of the problem, and do so in a way thatís helpful for the calculation. All right. So for us actually, although it may not be immediately Ė it may not be evidence from what Iíve said, when you pass lines through the region, certainly in applications, when youíre sending X-rays through, you tend to send X-rays through along parallel lines. So for us, a natural configuration of lines in the problem is something like this. Hereís the region, and you may wanna pass through a bunch of Ė you donít send a single X-ray through, but you send a bunch of X-rays through along parallel lines, all making the same angle but parallel lines. And then you change the angle and send another bunch of parallel lines through, something that looks like this. Okay? So a family of parallel lines.

And is there a coordinate description of the family of all lines that makes it easy to write down lines like that? Well, yes. And itís actually Ė if you phrase it this way, itís not so unnatural. As a matter of fact, Iíd like to think itís natural, although everything looks a little unnatural when you first see it. What is common to these lines? The thing thatís common to these lines is they all have the same normal vector, or they make the same angle with the horizontal axis, say something like that. So thatís gonna be one of our parameters is the parameter that describes the orientation of the normal vector. As a matter of fact, letís start with a single line and describe a single line in a way thatís gonna allow me to easily describe families of lines like this. So for a single line, I could take something like Ė as a matter of fact, let me take a single line through the origin. As a matter of fact, let me make it look like it goes through the origin. Make a single line through the origin, all right? Like so. And that line is determined by its normal vector, or what is the same thing? Itís determined by the angle that the normal vector Ė Iíll call it phi Ė that the normal vector makes with the X1 axis. So I consider I fixed the two axes X1 and X2, and then the lineís determined by phi.

The normal vector which Iím gonna need, so let me right it down Ė as a matter of fact, Iíll take the unit normal vector is just cosine of phi sine of phi. Thatís a vector of length one thatís perpendicular to the line. The unit normal vector is say N is cosine of phi sine of phi, but I donít need two numbers. I donít need the cosine and the sine to describe it. I just need the angle phi. That describes it. And here, what is the range of phi? Phi should vary from zero to pi, and you strip the inequality at least on one of the sides, letís say on the right-hand side. Phi equals zero means the line is vertical. It means itís making an angle of zero with the horizontal axis, so thatís straight up. And then as phi moves, the line rotates, and when phi is equal to zero itís vertical. When phi is equal to pi, itís also vertical, but you donít wanna have the same line described twice, so you have a strict inequality on this side. So phi equals zero describes a vertical line. Phi equals pi also describes a vertical line. You donít wanna describe the same line twice, so you have a strict inequality here.

So thatís a single parameter that describes all lines through the origin. How do I get other lines? Well, I take a line with the same normal vector Ė and again, Iím interested actually Ė I have in mind later on Iím interested in families of parallel lines, so I take a line with the same normal vector, and the other parameter that determines it uniquely is its distance from the origin, so a line not through the origin. The picture would be something like this. It has the same Ė hereís the line through the origin with that normal vector. So hereís the same normal vector. Hereís the angle phi. But thereís also a distance rho from the origin. Iíll call it rho. So itís determined by phi and a distance rho from the origin, and that determines the line almost. Iíll say a little bit more here. Thereís a slight subtle point here because I donít just wanna take the distance. How do I distinguish between that line Ė we draw two lines, and I wanna be able to distinguish between the two. Hereís one line, and hereís a parallel line. Theyíre the same distance from the origin. Hereís a normal vector, and hereís a normal vector. The normal vectorís unambiguous because I decide that phi goes with Ė the direction of the normal vectorís unambiguous because I decide goes between zero and pi, so if the normal vector Ė if this line goes that way, the normal vector for this line also goes that way. Theyíre both the same distance from the origin. How do I distinguish between the two? I donít consider just the distance, but I consider the sine to distance. So I consider the sine distance.

So what I mean by that is rho is positive if I get from the origin to the line in the direction of the normal vector, and rho is negative if I get from zero to the line in the direction opposite the normal vector. Rho is positive if I get from the origin to the line in the direction of the normal vector. So that would be the case here. The normal vectorís going that way. I get from the origin to the line by moving in the direction of the normal vector. Here rho is negative because I go the other way. Here rho is positive. Here rho is negative. So rho is just Ė Iíll write it down. Rho is negative. Rho is equal to zero of course if I have a line for the origin. So rho is negative if I move Ė if I get from origin to a line by moving in the direction opposite to the normal vector, like this case. So rho can vary between minus infinity and infinity, so the range of my parameters, the range of my coordinates for the lines is Ė and I get all lines this way Ė is zero less than or equal to phi less than pi and minus infinity less than or equal to Ė less than rho less than plus infinity. I describe all the lines in the plane by those two coordinates. So a give rho and a given phi tell me a given line, specify a given line.

And youíll notice that actually families of parallel lines are described very nicely in this coordinate system because a family of parallel lines means phi is fixed and rho varies. So a set of lines like this Ė something like that or at a different angle would be phi fixed, rho varies. Okay? Thatís nice. Now how do you write the equation of a line? I wanna write the Ė transforms are set up generally speaking Ė and weíll see actually. Weíre gonna bring the Fourier transform into this in a way thatís just unbelievable Ė in Cartesian form, so I wanna write a Cartesian equation of the line in these coordinates Ė of a particular line. So what is the Cartesian equation of the line specified by a given pair rho, phi? Notice here I have a distance and I have an angle, but these arenít polar coordinates. These arenít polar coordinates in the plane. These are line coordinates. These are coordinates of lines. Well, thatís not hard to see, and I will refer you to the notes for a little bit more of a derivation of this. Let me just say what the formula is, and again itís something not actually so different from what youíve seen before most likely, but maybe not written in quite the same terminology or quite the same language. Itís given by in vector form X dot N equals rho, very simple. N is equal to cosine phi sine phi. Thatís the Cartesian equation in vector form of a line which is given by a given normal direction, a given unit normal, and a distance rho from the origin, sine distance plus or minus.

And if I write that on coordinates, it would be cosine of phi times X1 plus sine of phi times X2 equals rho. Thatís the equation. Now thereís more magic. And again here, unfortunately I donít have time to derive this for you, and Iím gonna see Ė I have a set of notes on this, but it gets a little involved. Not so bad, itís quite interesting. Itís another example of how things in higher dimensions can be a little richer. The variety of objects you find at higher dimensions can be richer than in one dimension, and this has to do with evaluating a line integral via a delta function. So I wanna use this to evaluate an integral like this Ė an integral a function of two variables, X1, X2, along a line via a delta function. What you do is you consider Ė so here I just have to state a fact for you. And then with this, we are set to go, we are good to go for the rest of our trip. You consider Ė I hate that, like a mathematical weasel word, ďconsider.Ē Like what could be more natural in the world to consider the following ridiculously complicated looking object. Consider a delta function along the line. You can consider a delta function along the line described this way, so describe the way I wrote it. Let me write it like this: rho minus X1 cosine phi minus X2 sine phi equals zero. That is to say delta of rho minus those things, X1 cosine phi minus X2 sine phi.

Now you can think of this Ė so this is so-called line impulse. So you can think of this intuitively as a delta function which is concentrated along the line, so itís infinite Ė we donít think these ways because weíre very sophisticated. Itís a distribution, blah blah blah, but if you wanna think about it easily, think about it as concentrated along the line. So itís infinite along the line, zero off the line, and it has a special property with respect to integration. So itís infinite along the line, zero off the line, and the key property is for us is what it does under integration. Itís analogous to the one-dimensional case, but itís perfectly suited to our problem. So zero off the line like all good deltas, infinity on the line Ė I can write this shamelessly, right? And for the integral what happens is Ė Iíll give this a board of its own. This is really a key step because it allow Ė well, youíll see. One thing at a time. The line integral of mu along the line Ė so imagine mu is Ė when youíre integrating mu along the line, itís as if you want to concentrate mu along the line or just restrict mu to the line. Thatís given by integrating over the whole plane mu against the delta function concentrated on the line. This is mu of X1 X2 integrated against the delta function of the line or the line impulse associated with the Ė the line impulse for the line. X Ė rho minus X1 cosine phi minus X2 sine phi VX1 VX2. Now thatís a fact that Iím not gonna prove for you. You have to know a little bit more carefully how line impulses are defined and so on, but it is analogous to how delta functions work in connection with integration if you think of them in terms of integration with the one-dimensional case, or you integrate a function against the delta function Ė it consecrates the function at a point.

Here the delta function is concentrated along a line. To integrate a function against the delta function concentrated on a line, it sort of drops at one dimension so to speak and integrates the function along the line. You havenít seen this formula, but it is analogous to the kind of formulas you have seen before, and if you really want, Iíll post a set of notes on line impulses. I havenít had a chance to incorporate them into the notes because they need a little work, but itíll give you a rough idea if you really wanna know, but itís dangerous stuff. Watch this. We are set. Iím gonna invert the Radon transform. It is unbelievable. So again hereís the setup. Iím imagining myself Ė although you wonít necessarily see it in the calculation, Iím imagining that Iím fixing phi and Iím letting rho vary. You wanna use this to invert the Radon transform mu. So again, mu is this unknown function. I know the values of R of mu because I know the values of the line interval. Remember, R of mu Ė Iíll write it down one more time. R of mu Ė I could even write it in Ė R of mu in coordinates would be Ė say rho phi would be the integral of the line specified by rho and phi of mu. Itís the integral along the line. And I know how to concentrate that. I know I can write this out. I can write this as Ė that line integral can be written in turn as a double integral Ė let me write the whole thing out here.

The integral from minus infinity to infinity of mu X1 X2 delta rho minus X1 cosine phi minus X2 sine phi DX1 DX2. Great, so Iíve now replaced a simple looking formula by a complicated looking formula, and if you like that, youíre gonna love whatís gonna happen now. Just follow along for the ride. Enjoy it. Donít even take notes. Just sit back. Relax. The hostesses will be by soon to serve nuts and drinks. Now think about phi fixed to let rho vary. Think effectively about passing a family of parallel lines. So think of phi fixed and rho varying. So think about effectively computing this thing along a whole bunch of parallel lines. So youíre computing all these values as a function of rho. So with phi fixed and rho varying, you think about R mu rho phi as a function. Think of Ė as a function of rho with phi fixed. Now Iím going to before your very eyes take the one-dimensional Fourier transform of this function with respect to rho. Like, of course. Like what else would you do? I will take the 1-D Fourier transform of this Ė that function with rho varying Ė with respect to rho. In other words, I wanna computer the integral from minus Ė letís call it something like this. Letís call it the Fourier transform in the rho variable of R mu rho phi.

Now I need a dual variable, right? Because the Fourier transform always has a dual variable in the space domain and the frequency domain, so let me call it R. So let me write this as the integral from minus infinity to infinity of E to the minus two pi I R rho. Let me write it up here so itís Ė let me write it bigger. Phi just tags along for the ride. As a matter of fact, Iíd better write this a little bit more carefully already. Itís hard to write. Weíre gonna get caught a little bit here in problems of notation and problems of variables. Itís the usual thing in a subject that you know Ė there are problems here with variables. There are problems with naming your variables. Iím gonna take the Fourier transform. Phi is fixed. Rho is varying. Iím gonna take the Fourier transform with respect to Rho, and I need to say what the variable Ė I need to give the variable in the spatial domain a name, so Iím gonna call it R. So that would be like the integral from minus infinity to infinity E to the minus two pi I R rho, Radon transform of mu at rho and phi integrated with respect to rho. Then what pops out is a function of R and phi because phi just tags along for the ride. So phi is fixed, but thatís an additional parameter thatís entering into the definition here, but Iím integrating with respect to rho. So what pops out is a function of R. Think of it as a function of R, but then I say that phi is sort of tagging along for the ride.

All right. Now be not afraid. Write this out. Be not afraid. Be of good cheer. The holidays are close. What is this? This is the integral from minus infinity to infinity E to the minus two pi I R rho Ė oh, of this double integral minus infinity to infinity, minus infinity to infinity of mu X1 X2 delta X Ė delta rho minus X1 cosine of phi minus X2 sine of phi DX1 DX2, and then Ė thatís the R. And then the whole thing is integrated with respect to rho. Thatís just rewriting this equation substituting what I had before over there writing the line integral of mu in terms of the integral against the line impulse, the integral against the delta function. Okay? Be not afraid. Be not afraid. Flip the integrals. Do all sorts of unnatural acts. Write this as the integral from minus infinity to infinity, minus infinity to infinity mu of X1 X2 Ė I wanna bring the other stuff inside Ė the integral from minus infinity to infinity of E to the minus two pi I R rho times delta of rho minus X1 cosine phi minus X2 sine phi. That inside stuff is integrated with respect to rho, and then the whole thing is integrated with respect to X1 and X2.

So all I did was I sort of flipped the order of integration here. Inside here, first I was integrating with respect to X1 and X2, and then integrating with respect to rho, and there was this complex exponential outside. Now Iím flipping this thing around. Iím bringing this mu X1 X2 outside. Iím putting this with that. Iím integrating with respect to rho, and then the whole thing gets integrated with respect to X1 and X2. Be not afraid. Now look at that inside integral. Weíve seen integrals like that before. We have. Think about this as an engineer, not as a damn mathematician. As an engineer youíre integrating a function against the delta function. Integrating a function against the delta function evaluates the function where the delta Ė itís the shifted delta function. Youíre integrating a complex exponential with respect to rho against the shifted delta function rho minus Ė hereís the shift. All right? Hereís the shift. So what is that integral on the inside? The integral from minus infinity to infinity of E to the minus two pi I R rho delta rho minus X1 cosine phi minus X2 sine phi D rho is E to the minus two pi I R X1 cosine phi minus X Ė or plus X2 sine phi. Thatís it. Youíre integrating a function in a complex exponential against the delta function integral phi of X times delta Ė integral of F of X times delta X minus Y is F of Y. You know what Iím talking about here. Itís the integral. Itís the convolution. Itís the integral of the function against the shifted delta function. So that substitutes Ė the shift, itís rho minus Ė let me write it like this. Rho minus this, so it substitutes that into the exponent. Okay?

Let me write that out a little bit more. This is E to the minus two pi I R Ė let me write this X1 times R cosine phi plus X2 times R sine phi. And now, Iím gonna introduce some new variables here. Iím gonna introduce sort of dual variables to X1 and X2. So introduce C1 is equal to R cosine phi. C2 is equal to R sine phi. Now these are not polar coordinates on the C1 C2 plane, but you can sort of think of them that way. [Inaudible] because R is the dual variable to rho when youíre doing Fourier transform, and phi is an angle in the X1 X2 plane, so this is sort of Ė these are not exactly introducing polar coordinates in the C1 C2 plane, although there is an interpretation of them that way. But if I do that then that exponential becomes E to the minus two pi I X1 times C1 plus X2 times C2. Now all this was evaluating this inside integral here, the inside integral here. The integral of E to the minus two pi I R rho times this line impulse. And that emerges to be this complex exponential E to the minus two pi I X1 times C1 plus X2 times C2 when I make this change of variable.

Plug this in to my formula. Plug this in. Plug this into what? Plug that into that. This says the Fourier transform in rho of R Ė this Radon transform of R mu rho phi is equal to the integral from minus infinity to infinity, the integral from minus infinity to infinity mu X1 X2 Ė thatís what was leftover here Ė times this integral which I just computed to be E to the minus two pi I X1 C1 plus X2 C2 DX1 DX2. What do you see? You see the two-dimensional Fourier transform of mu. This equals the Fourier transform of mu at C1 C2, 2-D Fourier transform. All right? Letís recap. Fix phi. Let rho vary. Take the 1-D Fourier transform of the corresponding Radon transform. That is take the one-dimensional Fourier transform of this function of rho phi with respect to rho. What does it produce? It produces the two-dimensional Fourier transform of mu. Now in principle, the problem is solved. You know this. You know this expression. You measure these values. This is what youíre measuring. Youíre measuring the Radon transform, thatís the line integral of mu along this whole family of lines. As rho is varying and phi is fixed, you are measuring these values. You know this function.

Because you know this function, you can compute in theory its Fourier transform. You can compute its Fourier transform with respect to rho. Now what does that tell you? Computing the Fourier transform of this thing with respect to rho gives you the two-dimensional Fourier transform of mu. That tells you you can find mu by taking the inverse two-dimensional Fourier transform of what you have found. You can now find mu, so you compute once again the Fourier transform in the rho variable of this R mu rho phi. And that gives you a function. Now you have to make the changes of variable and all this jazz. You have to substitute C1 equals R cosine phi and C2 equals R sine phi and all the rest of that stuff. Okay, but what that results in at the end of the day is a function of two variables, C1 and C2, so you make the change of variables. C1 is equal to R cosine of phi. Thatís known. C2 is equal to R sine of phi. Thatís known. So this thing results in a function of two variables.

Just wanna make sure you understand exactly why this can solve the problem and in what form. So you compute the one-dimensional Fourier transform of this. You make this change of variables. This results in a function Ė letís call it G Ė of C1 and C2. Again, you know this function. You can compute its Fourier transform. You make these changes of variables. That gives you some function that youíre calling G of C1 and C2, and you know that that function is nothing other than the two-dimensional Fourier transform of mu. So you know that G of C1 C2 is the two-dimensional Fourier transform of mu of C1 C2. And you have solved your problem because now you take the inverse two-dimensional Fourier transform of this function which you have computed, and that gives you mu. Now get mu is the inverse two-dimensional Fourier transform of G. Let me just write it like that. That recovers mu by knowing all those line integrals. Done. Amazing. Itís absolutely incredible. So this is the mathematical basis of CAT scans, of recovering the density function by knowing all the one-dimensional slices. Now again, I set this up this way Ė although the modality, although the imaging techniques in the sort of whole physics behind it is different for MR, at the end of the day actually for amazing reasons, surprising reasons I think, the same thing holds.

That is the idea of finding the Ė when you try to do MR imaging, youíre confronted with the same sort of problem and itís solved the same way. Ultimately, you compute a two-dimensional inverse Fourier transform of like a one-dimensional Fourier transform of some auxiliary function that you introduced. It comes about in a different way, but it amounts to the same thing. Itís just amazing. Now again, none of this was known when Radon introduced this idea way back in I think either the end of the 19th century or the beginning of the 20th century Ė maybe in the Ď20s, 1920s. Iím not sure when. I have to look it up again. And then found this absolutely stunning and incredibly important application. I think itís nothing short of amazing. I think itís just absolutely amazing. Now of course, to make this practical and to implement this requires a lot of work. Everything has to be done discretely. If youíre actually gonna calculate this numerically, you have to implement discrete versions of this. There are a lot of computational issues. There are very interesting issues actually with the playoff between Cartesian and polar coordinate representations of the different quantities. There are lots of things that have to be done to actually make this practical, but this is the basis for it. This mathematical derivation, this way of going from the one-dimensional Fourier transform in the one variable to the two-dimensional Fourier transform in the other variables and allowing that to solve the problem is the basis for the entire thing. And I say I think itís just absolutely remarkable.

And I think with that we are done. So I wanna thank you very much for a stimulating quarter. I hope you enjoyed it as much as I did. I think this is wonderful material. I really do think itís every day another miracle almost. And Iím sure youíll find this material as you go on in whatever you direction you go on in your studies. Whether itís electrical engineering or other different fields, youíre gonna find the Fourier transform and its techniques used absolutely everywhere. And it has been my pleasure to give you what I hope has been a good introduction to it, so thank you very much.

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Duration: 50 minutes