 There is only one sign, please, to do. Let's restart, and continue with, say, Miguel Lectros, about three of these two. Of course, Miguel will be the streamer today, but I have to present you. I cannot share myself, yes. You can start. OK, thank you. All right, well, good morning. Thank you for being on time. In this case, let me turn the lights on here. That's not good. It will take me a few tries. So yesterday we were discussing Fourier transformations, and the last thing that I talked about is why they are so common in physics and in optics. And I was talking about these systems that might have satisfied two properties, at least approximately. They might be linear and shift invariant. So if I have a system, I send some signal of x, or of time, or whatever the variable is, and I get a response. Again, I'm going to use this shorthand, just so I don't draw that, make this system box very small. So I don't have to draw the big box all the time. And then the system is linear if for a signal 1 plus some amount of it plus some amount of a signal 2, the response is what? The same thing, but with responses. So each signal has its own response. Each one is scaled in the same way. And if I add 2, I get the sum of 2. Then that's what makes the system linear. What makes this system shift invariant is that if I send the signal with some delay, some shift in other direction, this gives me the response also shifted. So this is linear and shift invariant. As I mentioned last time, lots of systems are approximately linear and approximately shifting there. Suppose that the system has this property. What about if I send, so I know that if I send this signal, I get this response. But what if I decide to send the derivative of the signal? Do I get the derivative of the response? Let's think about it. What is this here? How do I define the, let me put question mark. Yes, so let's break it into pieces. So you're right, but let's see what that means. So how do we define the derivative? We define it as the limit. Sum delta x goes to 0 of signal of x plus delta x minus sigma of x, all divided by delta x. So now this is the sum of two terms. One is negative. They're both scaled by delta x. And this one is shifted. So it's not only that it's linear, a linear operation, it's also we need the shifting variance as well. Otherwise it wouldn't work. So we need it both to be linear and shift invariant. So this term then, well, the limit is just limit. But this, because of shifting variance, has a response that is just the shifted response divided by delta x. And then because of linearity, this other term is just also. And then we see that, yes, this is nothing but r prime of x. But this is true. And the same would be true for any derivative. So it is a unique property of linear shifting variance systems that any derivative goes to the same derivative. So now based on this idea, what if I send a signal derivative minus a constant? Let me say this would work for any constant. But I'm going to, for the sake of the argument, make it imaginary. And for sake of convenience, I'm going to put a 2 pi there. And then some other, for now, is a constant times s of x. If I send this combination of things into the system, what do I get? Well, the system's linear, so I get the response to this, which is. And then I get the response to this, which is a constant times this. So I just get minus i 2 pi nu r of x, sure. But now I tell the system, ah, I fooled you. Because I'm going to choose my function s so that this is 0. I can always choose a function s so that that is equal to 0. And 0 has to go to 0. What function does this? s of x is equal to what? Some exponential. Exponential of this stuff, i 2 pi nu, the variable, with some arbitrary amplitude. I know that 0 has to go in a linear shifting variable system to 0, which means that r itself is also a solution of this equation, which means that r is what type of function? It's the same exponential with one possible difference. Sorry? The amplitude. The amplitude, yeah. So I'm going to say this is some other amplitude, r0 e to the i 2 pi. So any exponential, imaginary exponential like this, goes to an exponential of the same form, maybe with a different amplitude. And for any nu, that is true. So exponentials play an important role in linear shifting variance systems. There's what sometimes we call the eigenfunctions. They are the functions that if you send into a linear shifting variance system, they come out the same way with a different amplitude, maybe. That's clear? So if I send, let me remove the amplitude just like this. If I say e to the i 2 pi nu x, and I send it through the system, I get e to the i 2 pi nu x times something. And I'm going to call that something h. And we don't know that this h is the same independently of what nu we have here. In the general case, it's not. We shouldn't assume that it is. So I'm going to say this h is therefore a function of nu. But it's independent of x. That's what is important. To the i's of the system, the variable is x. Nu is just a constant to the i's of the system. So an exponential goes to the same exponential with some scaling that might depend on the frequency. So this is starting to smell like Fourier. Is that clear? So what if I don't have an exponential? So for a general signal of x, well, in general, this is not an exponential. Can we write it as a combination of exponentials? Sure. That's what the Fourier synthesis is. So I can write this by using Fourier theorem as simply this amplitude e to the i 2 pi nu x d nu. I can always express any function that is not too badly behaved, that is square integrable, and has these properties as this proposition. And now if I send this through the system, what do I get out? So a sum goes to a sum because it's linear. What happens to this amplitude? This is a constant to the i's of the system. It's not a function of x. It's just a weighting of each one of the exponentials, so that stays the same. But what happens to this exponential? This one picks up an extra nu e to the i 2 pi nu x d nu. But this is equal to r of x, the response, which is equal to the integral for minus infinity of the Fourier transform e to the i 2 pi. And now we see what this system does. Fourier transform of the response is what? If I compare this to this, they're identical except for what? Instead of this, I have Fourier transform of the signal times this thing. So any linear shift invariant system doesn't matter if it's a circuit, a mathematical formula, a beam propagating in space, a camera, a music, like a stereo music system. If approximately it is linear and approximately it's shift invariant, approximately what it does is it just takes the signal, the Fourier transform, filters it in some way, multiplies it by something, and then gets the inverse Fourier transform. So the Fourier transform variable has a meaning because that's the space, the abstract space in which we're doing the filtering, where we just multiply. And that gives it some physical meaning. So that's why Fourier transformers are so useful, because this type of system, linear shift invariant, is so prevalent that the Fourier domain is going to be very useful. Any question? Yes, absolutely. So I was lazy and did it for one, but there's nothing in what I did that would not be valid for a multivariable system. And in fact, when we talk about imaging, we're going to talk about x and y, for example. But all the steps that I did would be valid, because linear shift invariant, that's independent of the variables. Then with shift, we would need to be invariant to shifts like this, let's say, and shifts like this, linearities independent of number of variables. You could do this combination of the derivative and the function for the derivative in any of the two directions or any combination that would give you an arbitrary exponential, say, in new x times x plus new y times y. And that would be what you would use in a two-dimensional Fourier transform, or three-dimensional or any dimension. So the same ideas would hold no matter what the dimensionality. So this thing is called what? This function that multiplies things. Anyone knows the name? It's called a transfer function. And it fully defines the system. Linear shift invariant system. So full description. So if I have a signal, go through the system, I get a response. Another way to think about it is I go this way. If I'm going to do this operationally, I do a Fourier transform that tells me to s tilde of new. So it's a Fourier transform that goes from x to variable new. Then what the system does is just multiplies this by this transfer function, does a filtering in Fourier space. It just multiplies by a filter that depends on new. And this can be a complex function. And then this last step is what? So a Fourier transform, I multiply, and I have to inverse Fourier transform. And that's what a linear shift invariant system does. When we talk about propagation light in free space or modeling an imaging system, we're going to come up with this type of thing. So both free space has a transfer function. An optical system has a transfer function, et cetera. There's one part of a connection we're still missing. How to go from here to here directly without going down this way? Sometimes you want to do it that way. Sometimes you don't. Doesn't matter. It's the same. But we need to do some property of Fourier transforms before we discuss this other path. All right, so by remembering these formulas for the Fourier transform, so if you have this in your notes, now I'm going to give you a short exercise because it's going to be useful for what I'm going to do next. So let me find this. So can you please calculate the Fourier transform of only 1, 3, and 6? The others we'll do later. So do the Fourier transform of delta function, of a rect function, and of a Gaussian. And I'll give you a few minutes. And if you don't finish, that's all right. I just want to use them for what I'm going to do next. Remember the formula for f tilde of nu is integral from minus infinity to infinity f of x to the minus i to pi. Do you want the curtains open or close? Doesn't matter. Are you registered in the school? That's OK. I think you've arrived a little late. And there's a list that we've been signing every day. Have you signed it? Ah, perfect. Is this one easy or hard? Don't worry. How many people have finished the first one? Let me do the first one here. So the Fourier transform of the delta function is just the integral from minus infinity to infinity of delta of x e to the minus i to pi nu x. What's this equal to? So what's good about having a delta function inside an integral? Just removes the integral. So this is equal to? Sorry? It's constant, because I can think this is x minus 0. So then whenever I see an x, I replace it with a 0 and I remove the integral. e to the 0 is 1, 1. So the Fourier transform of a delta function is 1. And again, this is something that would make mathematicians nervous, because this is not the type of function that is supposed to have a Fourier transform, because its modular square is not integral. But we physicists don't care much about that. We know what we mean. And this makes sense, because then the inverse Fourier transform, let's say calculate the inverse Fourier transform of this, is precisely the formula that we gave for the delta function. If I have the integral minus infinity to infinity of 1 e to the i 2 pi nu x v nu, this is exactly what we just did. But 1 here doesn't matter. This is delta. So it makes sense. I'll skip the rect, and I'll go to the Gaussian. So if I have a Gaussian, which is going to do a simple Gaussian like this, e to the minus pi x squared, then g tilde of nu is the integral from minus infinity to infinity, e to the minus pi x squared, e to the minus i 2 pi nu x dx. And what do I do now? I can combine these as e to the minus pi x squared. And I'm going to do some factorization. So plus i, I leave some room dx. And why did I leave some room? Same trick as yesterday. This is a square of something plus 2 times something times that something. It's asking to be so I can put i nu squared here. But I don't want to, I cannot just do that. So I have to cancel that by putting up e to the pi with the opposite sign here, i nu squared. And this whole thing is just x minus, not plus i nu all squared, minus infinity to infinity. And I can call this x prime. So this is equal to the integral of e to the minus pi x prime squared and dx and dx prime are related how? The same because just a shift. And again, this is from minus infinity plus a little imaginary part to infinity plus a little imaginary part. I'm not going to worry much about those imaginary parts, small imaginary parts compared to infinity times e to the, what is i squared minus pi nu squared? And what is this thing here? The integral of the Gaussian. So do we remember our little formula? So you say it's 1 because it's 2 pi divided by, if I had a 2, what is here? Well, that's a 2 pi. So for this particular one is 1. So the result for this one here, the Fourier transform of a Gaussian is a Gaussian. And that's a very important result because Gaussians show up everywhere in physics and in optics. And if we use this convention of the Fourier transform and this type of Gaussian, then it's identical, we just change the variables. I should warn you, I did not warn you, but should warn you with the Fourier transforms, everyone likes their own form of the Fourier transforms. Depending on what course I'm teaching, I use a different form. Here I'm saying nu x with a minus sign here to go into Fourier transforms. Some people like using plus and then minus for the inverse. Some people don't like putting the 2 pi here. And then you have to put a normalization in front. And there's all possible combinations of these things. So you just have to be aware that if you see something that's slightly different, well, it's because for whatever reason, for that physical context, people do something different. But it's nothing fundamental. It's just a small difference. Let me just store here my results. So the Fourier transform as x goes to nu of delta of x equals 1. And the Fourier transform as x goes to nu of e to the minus pi x squared is e to the minus pi nu squared. So I didn't want to erase that yet. So the next thing we're going to do is discuss the properties of the Fourier transform. We use it so much that it turns out that to solve, for example, once we have solved this case, if we shift this function, it turns out we don't have to do it again. There's a simple trick that lets us do it, or like here. If I shift this rect given that I know this one, I know what I should do here. Also, if I scale it, so if I take the variable and divide it by something, I don't have to do it again. There's a trick for how to do it, the scaling, once I know this one. Or if I know the Fourier transform of two functions, in this case the same one, I can know the Fourier transform of their convolution, or et cetera. Or multiplication by x. So there's a lot of properties that once we know a few cases, then these other would be simple. You probably know those, so I'm going to go a bit quicker here. I'm just going to show them on the screen. And the details of the derivations are there. In the interest of time, we will only look at them quickly, especially the last property, which is very important, but I'll discuss in more detail. All right, so I'm almost there. There we go, properties. So the first one is, first of all, plancheurial theorem. Suppose that I have a function, and for example, in optics, if I have a beam going in this direction, and I look at the field, a given component of the electric field at a given plane, can I measure, if I put a detector here at CCD, can I see the field? Is what I see the field? No, what is it that I see? The power, the intensity, which is related how to the field? It's the modulus squared. So if I'm using a complex representation of the field and it take the modulus squared, that's what is directly detectable. So this is a property of wave fields in general, that the modulus squared is something that is easy to observe. In quantum mechanics, something like that happens. When I have a quantum wave function, which tells me something about the evolution of a particle or something, is the wave function a probability? No, the wave function is not a probability. I have to mod square the wave function, and that gives me a probability density, something that I have to integrate to get a probability. So the mod square of something that satisfies a wave equation is very important. It's something we observe. So if I have x versus f, typically what I observe is this. And this is always going to be non-negative. And it has to do with the power. And the fact that I said that Fourier transverse really should only be used for functions where this thing can be integrated means that I'm dealing with quantities that have a finite amount of power. This property then tells me something interesting. So I start with the integral of the mod square. Well, the mod square is, of course, the function times its complex conjugate. Now I can take the function and say I can write it in terms of the inverse Fourier transform. Is this too small? Should I make it bigger? I can write it in terms of the inverse Fourier transform. I substitute that here. Then what is the next step usually when I substitute an integral inside of another integral? You reverse the order. So I reverse. I pull out the integral in u. I pull inside the integral in x. And then I notice, ah, wait a minute. This is a star here. And this looks like the inverse Fourier transform, but it's missing a minus here. Well, I can fix those two things by putting brackets factoring out a star. And now there's some minus. And then I say, oh, this is a Fourier transform. Once I do that, then I have f tilde times f tilde star. That's the mod square of f tilde. And I get this nice property. That's the integral of the area under this. If I do the same thing with the Fourier transform, so I have f tilde. And again, f tilde directly is not observable, but f tilde mod square is observable. Let's say this looks very different. But what is the same? The area. This area has to be the same as this area. And this extends to any number of dimensions as well. All these, or several of these properties going to have some nice physical interpretations. And to do that, let me talk about one of many applications of Fourier transforms in optics. If I have a field, some field distribution here, and I want to find the Fourier transform here, what optical element can I use that does something like a Fourier transform? A lens. So many of you know that a lens does something very similar to a Fourier transform. So I can put a lens. I assume the lens is infinite. It's thin. So all sorts of approximations. This is just to give us some intuition. It's not the rigorous truth. It's just to give us some intuition. This distance here and this distance here have to be what? The focal distance. So this should be f, and this should be f for this lens. So this distribution then gives me this distribution. Sorry, this is, let me make this f different because it's not the function, it's the focal distance. It's not this f. So if I have f here, I have f. So what is parsable theorem telling me physically here? What is the physical meaning in this context? The conservation of Fourier. All the power that I have here is going here because I'm using the approximation that my lens is completely transparent, is infinity in extent, and no light is escaping it. So it's not filtering out any amount of light. All the light from here traveling in these directions making it here. So there's no power loss. The power here has to be the same as the power here. So simple interpretation. By the way, this has a nice interpretation before I keep going with the next property. And it's the first case in which we can sort of connect the ray optics, the geometrical optics picture of what a lens does with the wave optics picture of what a lens does. What happens if the field here is a delta function? What would that mean? Suppose that my field at this plane is a delta function. What does that mean? It's just a very bright point. That means there's no light anywhere. Lot of light here, no light here. So if the distribution of light here is a delta function, that means dark, dark, dark, dark, really bright, dark, dark, dark. So I have a lot of light here. What would the rays of light do there? They emanate from that point. And what do the rays do after that? They go like this. So what's their face here? The face of one ray versus the other. It's the same because they're parallel. So the field here is how? It's constant. It's the same field. This is distributing nicely all the energy into all these points. And not only I only drew these ones, but again, this is an idealization. This lens is infinite. So all possible points are illuminated. So the Fourier transform of a delta function is a constant, which in this unit is 1. So this is consistent with this picture. Delta function goes to 1. What if I had 1 here? What would be my Fourier transform? 1 point, the delta function, because it's the same thing. So yes, sorry? The divergent lens is far more complicated for this analogy because it does not do a Fourier transform in a trivial way. There is a hidden Fourier transform in terms of virtual images. This is easier to see. So if you wanted to really do a Fourier transform, it's better to use a convergent lens. It has to be just the L2 norm with uniform weight. That's a property of Fourier transforms. For example, if I put here 1, then it's not valid. We can show that it's not valid. Or 3, or any number other than 2. This one means that if I have a lens here and I have a Gaussian distribution, how would I create a Gaussian distribution, for example? What optical device would give me something that is more or less Gaussian? Do I have one here? Laser pointer, for example. Although that's very wide. But if I were to focus it down, it would give me a small Gaussian here. And notice that this is not any Gaussian. It's a very specific Gaussian. So it's a very specific Gaussian that is Gaussian here. And this is going to expand. And then this lens is going to refocus it into a Gaussian of the same size. But it has to be the right size. That's why this is not true for any Gaussian. It's only for this particular Gaussian. So that's the first property. The second property is the shift phase property, so-called. So suppose that I know the Fourier transform of f. What if I now calculate the Fourier transform of f minus x naught? I shift my function. Well, for all these properties, what you do is you substitute this shifted function, say. Then you do a change of variables. I call whatever is inside that a new variable x prime. So I do that change of variables. So now this f of x prime. But this x now becomes x0 plus x prime plus x0. The x0 part I can pull out of the integral. And this is now the Fourier transform of f. So I have the Fourier transform of f times a linear phase. So if I know the Fourier transform of f of x, then shifting the function just causes a phase of the result. And let's see what that means over there. And I think I have colors, which is good. For example, for the delta function, what if I shift my delta function? So now the bright part is here. What would the rays do in this case? They go in all directions again. And then in what direction is it going to go? Which ray is easiest to see here? The one that goes through the center of the lens, as Somberto was telling us. That one, what happens to it? Goes straight. And all the other ones go in what direction? Parallel, because we're in the back focal plane. So they all go like this. So the rays are nicely equally spaced again. So we also have something with constant amplitude. But because the rays are going at an angle, that means that the wave fronts are tilted. And the phase increases or decreases in this case as we go down. That is, we're multiplying the answer by a linear phase, which is exactly what this tells us. And the more we shift it, the more rapidly that linear phase oscillates. So again, this property has a nice geometrical meaning over there. The same would happen here. If I take my Gaussian and I did this, it would get focused down here. It would be the same Gaussian. But because it's tilted now, it has a linear phase. Clear? So we do that. There's the opposite rule, which is, if I know the Fourier transform of a function, now what happens if I have function times a linear phase? And I take the Fourier transform of that. You can just substitute that mathematically and you find that it's simply the shifted Fourier transform. So what would that correspond to in this case? If I have, let's say, the opposite case where I have the parallel rays and they're going at a given direction, it's going to shift my delta function on here. Or if I have this beam and it's like this, but I have a linear phase, which means that it's going to be propagating in some direction, then it's going to shift it here. So a linear phase on one side is a shift in the other and the other way around. Now, what about scaling? So suppose that I have the Fourier transform. I know what it is. Now I say f of x divided by a. So if f is this, what is f divided by a? Let's say a equals 2. Is it wider or narrower? Has to be wider because the argument has to be twice as big to cancel that 2. So if this is f of x, then this one here, let's say, would be f of x divided by 2, let's say. So this function gets wider. So if I take any f of x and I scale it, squeeze it or stretch it, then we confine again by substituting here, then I call this new thing inside x prime. I do a change of variables. Now x is a x prime, so dx is a dx prime. And I do the change of variables and this a comes out of the integral like this and it comes in here. And what I get essentially is a factor of a times the Fourier transform of a times new. So if the a was downstairs here, it is upstairs here. So what does that tell us about the Fourier transform? If I stretch this function, then what happens to the Fourier transform? It also stretches. So if I stretch the original function, the Fourier transform squeezes it. What about this amplitude here? So what this tell me is if I start with something like this and I go to something like, let's say this, well, not that doesn't satisfy Parseval. I go something like this, then if I were to stretch this one, make it bigger, then this has to go to something that is narrower and taller. Why taller? To preserve the norm because of Parseval's theorem. The area over this and this has to be the same if these are the norm squares. So if you stretch something, you squeeze the opposite. What would that tell us in these examples here? So I had my magic Gaussian here that is expanding and then refocus perfectly here. What if I make it wider? It goes narrower because this is closer to a collimated beam going in, getting all focused down. So if I have a wider Gaussian here, I have a more localized and taller thing. And of course, this is clear because we know that the more beam that we put in here and the more wide it is, the sharper focus we can achieve on this side. That is the first hint at something to do with resolution. We want sharp resolution. We want this to be very thin and intense, which means that whatever we have in the Fourier space has to be wide. Now this one is not so easy to see in this picture, but it's very useful. It turns out that if I know the function and it's Fourier transform, then the derivative is easy to calculate because I can just substitute the derivative and then there's this trick that you learn at some point when you were in university, which is called integration by parts, which lets you remove derivatives from something instead of an integral, one part and give it to another part. If we use that trick of integration by parts, I won't go through the details. You find that the Fourier transform of the derivative is just the Fourier transform of the function times I2 pi nu. So this is very nice because by doing a Fourier transform you go from taking a derivative, which is a harder operation, to just multiplying by the variable and a constant. And you can do this any number of times if I have the Fourier transform of the nth derivative of function, this is just this constant I2 pi times nu all to the n times the function. This is very useful when solving differential equations, some types of differential equations because you have an expression with different types of derivatives of the function. And if all the coefficients are constant and the range of integration is minus infinity to infinity, you can do Fourier transform of the whole equation and all your derivatives become multiplication, then you can solve this algebraically and then you just take an inverse Fourier transform. So it's really convenient. Now just for the sake of, just because it's fun, might not be useful, but it's fun, I'm gonna mention the following idea. So suppose that you want to take the derivative of a function in the computer. There's an easy way we will see later to do Fourier transforms once you have a numerical function. So a way to do the derivative is then how? I can take the Fourier transform, multiply by this an inverse Fourier transform. If I want to do the second derivative, two derivatives, I can take the Fourier transform, multiply by this squared and then they take the inverse Fourier transform. So I can do the nth derivative is the inverse Fourier transform of I two pi new to the n times the Fourier transform. So this is a way to do the derivatives in a way in the computer. Can I calculate the half derivative or the three quarters derivative? Why not? And here we always think of this as an integer, but here, hey, whatever, I can put whatever number I want here. So this gives me a way to evaluate what is called fractional derivatives. Something that is the half derivative of the three quarters derivative, the five eighths derivative of a function. Now there are some people that claim that these fractional derivatives are useful. I'm yet to be convinced that they are, but certainly this is the way how you would calculate. And looking at the derivative this way lets you do this. It's not just something that you have to do an integer number of times, you can do a fractional time. And I see that if I do the half derivative and then I take another half derivative, I get a full derivative. So it's just a, yeah, that's right, absolutely. So if you do negative n, then you have like a primitive, like an indefinite derivative, yeah. There are ambiguities in all these cases. If you do the negative n, it's the integral, but we know that those have an arbitrary constant that is undefined. If it is the first, the minus one derivative, if it is the minus two derivative, there are two constants that are undefined. And likewise, if this were a half, then essentially I'm multiplying this by what? So if this is one half, this is the square root. When you do the square root of something like this that is complex, there's an uncertainty in a sign. There's something that is called a branch point that the square root has two solutions. And when new is positive, you know what it is, but when new is negative, you're no longer sure what it should be. And then there's an ambiguity in the definition. So that's the catch. But for n integer and positive, there's no ambiguity. It's all clear, which is good because that's the usual derivative. The counterpart of this property is if I know the Fourier transform of f and I now want the Fourier transform of x times x to the n, that ends up being the nth derivative of the Fourier transform divided by minus i to pi to the n. So it is similar that's going from a, now if you multiply, it goes to a derivative. So this is not as usual. This makes your life harder in a way. That is a property. This one is extremely useful and extremely relevant to something we're doing here. We talked about this convolution yesterday, this blurring of a function by another function. What is the Fourier transform of the convolution of two functions? Well, I just substitute the convolution inside of the formula for the Fourier transform. I change order of integration. So I pull out the f from one of the integrals. Then I recognize this is the Fourier transform of shifted g. So this is just the Fourier transform of g times a linear phase. I put that in, reorganize. This is the Fourier transform of f. And what I find is that this is just the product of the Fourier transforms. So the Fourier transform of the convolution of two functions is the product of the Fourier transforms. Remember that we showed that contrary to our intuition, convolution is commutative. You can f star g is the same as g star f. In terms of the Fourier transform, now it's trivial to see it as a case because once we go to the Fourier transform, we have the product f tilde times g tilde. That, of course, we can reverse. So it is clear that this is true. Now, there's a very, very important consequence of this. So if I have two functions, I calculate the Fourier transforms and I multiply them and then I have to take the inverse Fourier transform. Of that, I get the convolution. When did we see the product of two Fourier transforms recently? Here. So we have the transfer function, which is something in Fourier transform space times the Fourier transform of the sigma. So the response is the product of two things. So if I define something called, let me call it g of x as the inverse Fourier transform as nu goes to x of h of nu, then what is the relation between r of x and the signal? And the signal is the inverse Fourier transform of this, which is the same as what? The convolution of signal with g. So s star. This thing here is called the impulse, the property, the duration of the property or how that connects to this? Okay, so we know that if I have the Fourier transform of f star g is f tilde star. So it's the same thing as saying, if I take the inverse Fourier transform now, f star g is the inverse Fourier transform of f tilde g tilde. What I have here precisely is that the response is the inverse Fourier transform of the Fourier transform of r of this. So if I take the inverse Fourier transform of this, I get the response. But this is this, so I just wrote s tilde times h. But that property tells me the inverse Fourier transform of a product is a convolution. So I have the convolution of the inverse Fourier transform of this with the inverse Fourier transform of this. So what that tells me is that the way directly to go here is saying that this is a convolution, a convolution with this thing called the impulse response. In other words, any shift invariant linear system blurs. That's all it does. Whatever the system is, doesn't matter if it is an electronic circuit, if it is light propagation in free space, it's a camera, what it does, it blurs. And it has its very own characteristic blur, which is its impulse response. It's a property of the system. So what the system does is it has its blur that it likes and whatever you give it, it blurs it with that. Which is the same thing as takes the Fourier transform multiplies by its transfer function, inverse Fourier transform. It's the same thing. So it's always a blurring. That's why this convolution is so important because all these systems, all they do is a convolution with their characteristic impulse response. They blur. So always a blurring. Why is this called the impulse response? The response, yeah, an impulse, a unit impulse is another name for the delta function. If my signal were a delta function, so I see the response to one thing, then what is the Fourier transform of the delta function? One, so this would be one times this and then I would get that the response is just G. So it's the response to an impulse, to a delta function. So in the case of a camera, if we ignore aberrations, which are what will break, some aberrations is what breaks the shifting variance. Then if I take a photo of a star, which is the closest thing we have to a delta function in X and Y is a very sharp bright spot. What we see is not just a star, we see a blur. And if our camera is good, it's a small blur. If our camera is bad, it's a big blur. That's the impulse response. That tells you, whatever I take a photo of, if I take a photo now of you guys, it's gonna blur you all with this same response that I would measure by taking a photo of a star. So any linear shifting variance system, all it does is blurring with its own characteristic blur. So when we do later today the physical optics lecture, well, I don't know, we'll get to that tonight or tomorrow, we'll see what is the transfer function and the impulse response of free space, simply a free space propagation. So that's fairly clear. Okay, let's keep going down. And this one also has its counterpart. If I take the Fourier transform of a product, I get the convolution in Fourier space. So it goes both ways. The proof is similar. All right, so this gives us good, because I wanna spend some time on this one. So most spokes on Fourier transforms finish here. They might have one or two more. But there's another property that I'm gonna spend some time discussing that, unfortunately, many courses or many books skip and that students learn, if there are physics students in a quantum mechanics course and then there's this misconception that this is a property of quantum mechanics and it's a property of Fourier transforms. And that has to do with this space bank with product or the uncertainty relation, which turns out to be the uncertainty principle of quantum mechanics. So the derivation, I'm gonna skip. You have access to this notes. It goes for one, two, three, four pages. But what is important is that when you do this in a quantum mechanics course, if you're a physics student, by then your professors have introduced all this notation of the operators and the commutators and the brass and the kets and all this. And you do it with that in a very elegant and compact way, but we tend to think, oh, this is quantum. This is inherently quantum. What these operators and commutators and all these things are are shorthand. It's linear algebra. It's not quantum. It's a shorthand that was invented by quantum people to describe these things, but you can do classical optics, acoustics, whatever, using the same notation. It's just a way of writing things in a short way. As I like to tell students, the fact that you see an operator and a bracket or a ket or a commutator doesn't mean that what you're looking is quantum. The people that invented quantum mechanics, many spoke German. That doesn't mean that when you hear German, they're talking about quantum mechanics. It's just the language they're using to describe quantum mechanics. So it's the same thing with the mathematical language because they're using that mathematical language is not that it's inherently quantum. So just to make that point, all the derivation that I skipped that I'm gonna leave you as homework to read is not using that language. It's just using regular calculus integrals. The equivalent of the commutator is a trick of integration by parts. It's 19 or 18th century mathematics. But the main point is the following. As I mentioned before, if I have some physical distribution, let's say F, what I observe is not, if this is an optical field, it's not the field itself, it's the modulus squared. That's like the power or the probability density or something. I have something, I don't know, something like this. And then in the Fourier space is the same thing. I cannot observe F tilde directly. Usually what I observe is the mod square of this and this might look somehow different. Let me make it something like that. Is my drawing correct or not? So what should be the same on both spaces? The area. Well, of course they have different units so I can always say that I'm scaling it, but sure, the area should be the same. That's what Parseval's here and tells me. Now, how would you distinguish something like this from, let's say, something like this? Oftentimes we're interested in the centroid of a distribution, also in the width of the distribution. How would I define the centroid? Oh, I give the mathematical formulas there, but it is like the center of mass. Imagine that this F squared determines the mass density of a linear distribution, like a stick, that is heavier at some bits than others, and then you look for the place where you can balance that stick on your finger and it doesn't go one way or the other. It's a nice demonstration. I don't know if this works. It's more dramatic with something asymmetric, but if, try this with a broom at home. So how do I find the center of this? Here is symmetric, it's very easy. I just go to the center, but if it were symmetric, you just do this very slowly and it always goes to the center of mass. Even if the object is very symmetric because as soon as you move something and it's heavier on this side, there's more friction on this side than this one, then this hand moves and it balances out. You always get to the center of mass. So that is what this center of mass here would be, the centroid. How would you estimate how wide this distribution is? And there are many ways. What measure tells you what that distribution is? The standard deviation is one and it's the one that we want. It's not the only one. In optics, we use a lot. Oftentimes people don't want to calculate that and they want to do something you can do with your eyes. That's what's called the full width half maximum, for example, you find the maximum, you come halfway and you see what the width is. That has a lot of problems, of course, because a function like this and a function like this have the same full width half maximum. And yet this one decays much slower. Why? Because all the difference happens below the half-way line. So the full width half maximum is imperfect. They're all imperfect as it turns out. There's another one that is related to something we use in optics that is called the Strel ratio, which is you calculate the area, you know the height. If it were a rectangle, the width is what? So if I have a rectangle and I know the height and I know the area, how do I calculate the width? Area divided by height. Well, you can use this for any function and it gives you a measure of how wide they are. That's another usual measure. But the one that you mentioned is the one we're gonna use, but it's important to be aware that this is a specific choice, is the standard deviation. And that has a more complicated formula in terms of an integral. It's the integral of x minus x bar squared f squared dx divided by a normalization. And this is the standard deviation squared, the variance. Or in other words, it's the square root of this whole thing. And that's what we would use, for example, if I were a teacher for a big class and these were the distribution of grades that I got in my course and I wanted to know what the average of the class was and what the standard deviation of the class was. That's what I would use. Or the approximation where these are sums because students are discretized. So this is what we're gonna use. So there's a delta x here, which tells me how wide this is. But there's also a delta new here that tells me how wide this is. This is actually sort of like a half width. So delta x and this is delta new. Defined in the same way, but in terms of the Fourier transform. What this uncertainty relation tells me after a long derivation is that delta x times delta new cannot be smaller when I multiply them than a given constant, which is in this case, one over four pi. It just happens that that's not possible. So what is that telling me? If I make this function thinner, that one has to be wider. And there's no function that can be thin here and thin there. It's also shown in the notes, I seem to recall, that the function that does best at being thin here and there with this particular measure, at the same time, is a Gaussian, our old friendly Gaussian. It's the one that does best. It's the one that for which this is equal. For all the other functions, it is greater than that. So again, if this course was one week longer, I would do all the math, but we don't have the time. So this is very interesting. This tells me that I cannot have a function in one space with a given width and in the other space with the other width. And this has all sorts of physical implications in different fields, depending on what we're doing. Yes, sorry? The, this one? The centroid? No, no, no, let me think about this. Yes, yes, the area, no, no, no, it doesn't. Because it's, for example, if I have something like this and then something like this, the centroid might be somewhere like this. There's the extra thing of how far it is from the centroid. That also counts. So here the area is much bigger on this side and this side. So again, if you had a stick with very light, but with a mass here and with another mass that is five times bigger here, you wouldn't have to put one of the masses on top of your fingers to have this balance. You would go somewhere in between and a ratio proportional to the ratio of the masses. Any other questions? All right, so, perfect. So let's think of what this does in different areas of physics. In optics, for example, I'm gonna come back to this later, but I know that if I have this lens here, what is this telling me? If I want something very thin here, I need something wide here. The thinner I want it there, the wider I want it here. And this has to do with resolution. Later when we see the meaning of Fourier transforms in space, we'll give more details to this. In quantum mechanics, a nu is, turns out the momentum, and because the momentum doesn't have the right units, we need to use a constant, which is, we need to divide by Planck's constant. So then we get delta x, delta nu goes to delta p, divided by Planck's constant, greater or equal than one over four pi, or I can put this one here. And then there's a shorthand, h divided by two pi is what? h bar divided by two. It's just Fourier properties because the Fourier transform of position is momentum. And what does this tell me? In quantum mechanics, this would be the probability density of where the particle is. It's quite likely that it is here. It's the least likely that it is here. I'm not sure, it's a probabilistic theory. It's very unlikely that it is here. So delta x gives me a measure of what? The region where the particle could be, the uncertainty. So the smaller delta x is, the more certain I am of where the particle is. The bigger delta x, the bigger the uncertainty. But the bigger the uncertainty in delta, or the smaller the uncertainty in delta x, in momentum, which tells me how certain I am about what the velocity of the particle is, the momentum, then this is less certain. The more certain, the less certain. And the name uncertainty comes from this very specific physical application in quantum mechanics. But it's nothing uncertain about this relation. It's a deterministic mathematical formula for Fourier transforms. It's just when we add the probabilistic element of quantum mechanics, and that these deltas become uncertainties. And then this is an uncertainty principle that is very profound. It tells you the more you know about the particle's position, the less you know about the particle's momentum. Okay, so is that clear? So let me finish with music. So normally I would like to have longer to do this, but I try to do it in eight minutes. So new in music means what? So if x, if x is time, so if I have delta t, delta new, squared and equal than one over four pi, what is new? Frequency. So say that I play a note, let's say in what in English is called an A, what in other languages is Allah. I have a given frequency here. So 220, 220 hertz. One of them. If I want to play the next A, what frequency should I go to? The next higher A. It's twice this. And if I want to go to the next A, I have to do what? Three? Now it turns out it's four. You have to double the frequencies to go to the next octave. It's a logarithmic scale. So the next A would be at four. So that means that the frequencies for the notes that are equally spaced in the keyboard are spaced logarithmically. And it turns out that to go from A to A, how many keys do I have to skip? At least in the modern Western scale. How many keys do I have to go from an A to the next A? It's 12, 12. How many frets do I need to skip on a guitar from an A to an A? It's 12. And in fact, the scaling here is very similar to the scaling of the frets of a guitar. Which means that if I want to, because there's 12 notes and there's a whole story of why 12 and not another number, which is very nice, but I'm gonna have to skip. What frequency is this one then? One note away, a semitone away. To go from here to here, I have to multiply by something. And then to go from here to here, I have to multiply by the same thing again. And to go from here all the way to here, I have to multiply by the same thing 12 times. What number, if I multiply by itself 12 times, give me two? And this trick question, because it's very simple. It's two to the one 12. So if I have a frequency and then I multiply that frequency by two to the one 12, that gives me the next note in the musical scale. So the spacing here, then between these two frequencies is two, one 12 minus one times the frequency. Why do I care about that? What does this have to do with uncertainty? Let me come here. So this scale looks like that. That's the natural scaling of the frequencies. And suppose that I play a note. And I'm gonna play a note here, one of these notes. And I hope this is not too loud. So that's an A. That would live here. But in that case, delta T tells me about what for a given note. So if I have in the time domain, the note, what does this delta T correspond to? The duration of the note, how long I play it for? So a long note has a long delta T. A short note has a short delta T. So it's the uncertainty in that is when the note happens. So if I have a finite note, that means that I have a finite delta T and delta nu is going to be also something, some number. It's not gonna be zero, it's gonna be something. So I cannot have just the delta here. I have a frequency distribution for my note here. If I make my note shorter, what happens to this? It gets wider. And if I keep making it shorter and shorter and shorter, what's gonna happen to this note? It's gonna start being so wide that I can no longer call it a note. It's invading the neighbors. It's not really just an A. It's a mixture of a bunch of things that goes to the next note. So I can calculate by using the uncertainty principle how short my note has to be before it is no longer a note. And I'm gonna skip the calculations, but I'll just show you here. So now what I'm gonna do is with that note, I'm gonna put a Gaussian envelope because I know that Gaussians are the best for this. So now it's the same notes. Instead of having this, I have this. So it comes and goes. I can make my note shorter. That's the same note, isn't it? I can make it half as long. That's the same note, it's just shorter. And when I write musical notation, I can make my note as short as I want. Now this calculation here, which I don't have time to do, but you can see the two to the 112, minus one, the four pi, et cetera, and the frequency, this tells me how short my notes will be before I start invading significantly the other notes. So it's about 0.01. So it's about five times smaller than I have it now. Let's see, let me make this 0.02, twice as big as the minimum uncertainty. So is this the same note? Yeah, more or less. If I go to 0.01, then I should be starting to get into trouble to see that I got into trouble. Let me go half as of that. And someone in Skype just came in. So this is definitely below that. So remember the note. And I'm gonna play the short note. Is that the note? It is not. I cannot call it a note. It's so wide in frequency that it's not a note. So this is the uncertainty principle for sound. It's just telling you that you cannot make a note arbitrarily short and still call it a note. Now notice that this expression is proportional to the basic frequency, where it's new. That means if I went to a higher frequency, because what matters is the ratio between you and you. So if I go a higher frequency, I have more space. So I can make my note shorter and still hear it as a note because the spacing gets bigger and bigger and bigger here. That means that I can afford to make notes shorter when there are higher frequencies. Most people play, most people are right-handed or left-handed? Right-handed. And when we play the piano, what hand can move faster? The right. And the right is playing what notes? The ones with the higher frequency. Because we can play higher frequencies and still follow this with our ears. When you picture a very fast and virtuoso, let's say guitar player, say an electric guitar player, and they're playing something really fast. Are they playing here or are they playing here? They're always playing here in the very high-pitched notes. Or when you, same with a violin player, when they're playing the really fast things, they go like this. We can hear better melodies that are fast if they're high-pitched. Bass players that I know that are stand-up bass, which is an instrument without frets. So it's one that you have to hit the right note and with your ear. Tell me that if you're really good and you play really fast, it doesn't matter if you hit the note no one notices. That's because we're at low frequencies and you don't have, the ears don't have the resolution because of this uncertainty relation to tell those notes. Let me see if this works. I haven't used this in a while. So I coded Toccata and Fugue, a little piece. So can you follow the melody? Now I'm gonna make it shorter. So I'm gonna play it. This was over four seconds. I'm gonna make it over two seconds. It's starting to be a bit not very clear. If I double the frequency, so if I go one octave up, that's very loud. But you can follow the melody, can you? Much better. So you can follow fast successions of notes much better when they're at high frequencies. That's fully consistent with the uncertainty principle picture. So the uncertainty relation, I should say. I like making the distinction between the uncertainty relation which is a mathematical property of Fourier transforms and the uncertainty principle which is a very physical, particular property of quantum mechanics. But they're both part of the same relation of Fourier transforms. Okay, so the thing that I didn't cover that I will start next time is just very quickly generalizing this to two dimensions because we will need it for resolution. But I hope that this sort of gives you some refresher on Fourier transforms and their properties. Yes? How this is related to Laplace transforms? Ah, Laplace transforms are good for solving equations with initial conditions. So it's when you only care about from zero to infinity. Fourier transforms when it's from minus infinity to infinity. It is easier for Laplace transforms to change the exponent, remove this i. And then, but Laplace transforms have the same properties that derivatives go to multiplication and that's why we use them to solve differential equations. So the Laplace transform is the daughter of the Fourier transform mathematically. You can derive the Laplace transform from the Fourier transform but it's for very specific applications. All right, so coffee break, we come back at 11.