Lecture 13 | The Fourier Transforms and its Applications

this presentation is delivered by the Stanford center for professional development nope okay big day today big day big day it's finally time to reap some of the benefits of the discussion rather general and I think safe too fair to say abstract discussion we've been having it'll still may still seem a little abstract to you today I understand that so just ride along all right and you'll see today some amazing formulas that come out of this effortlessly in finding the Fourier transform of some well-known functions things we're really going to have to use it's really just I don't know just a pocketful of miracles and the fun never stops so I want to talk today about the Fourier transform of a generalized function or a distribution also known as distribution Fourier transform of a distribution so let me remind you first of all what the setup is what goes into this to define a distribution you need a class of first of all class of test functions so the setup is you first need you first have to define a class of test functions or test signals that usually have particularly nice properties for the given problem at hand and it can vary from problem to problem for us for the Fourier transform it's the class of rapidly decreasing function so these typically have particularly nice properties sorry for not specifying that care terribly carefully but they come generally out of again sort of years of bitter experience with working with problems working a particular class of applications and trying to decide what the best functions are for the given class for free a transforms for the Fourier transform Fourier transform the class of test functions is the rapidly decreasing function so I won't write down the definition again but I'll remind you of the main properties in just a minute what we needed rapidly decreasing functions these are the functions which are infinitely differentiable and any derivative decreases faster than any power of X so they are just as nice as you could be rapidly decreasing functions all right that's what you need you need a class of test functions and then by definition a distribution or a generalized function is to use the most compact terminology it continuous linear functional on the set of test functions distribution also called a generalized function this term is probably not used generalized functions probably not used as much as it once was when the subject was new nowadays people just call it distribution since Schwarz is really unifying work on all this is so let me give you the shorthand notation for this they continuous linear functional on test functions alright so that means it satisfies two properties so one you say you're right the pairing if V is a test function and T is a distribution we will write most of the time but not all that and all always T fee with that sort of pairing for T operating on feet all right so if you want you can think of fee is some distribution of physical information temperature voltage current whatever and T is a way of measuring it all right so you measure something you get a number alright so T is a measuring device and fee is the thing you are measuring so T operates on fee in such a way that it satisfies the principle of superposition or that is to say it's linear and it's also continuous so T is linear meaning T of V 1 plus V 2 is the same thing as T operating on fee one all right up here T operating at T operating on fee 1 plus T operating on V 2 and the same features for scaling so T apply T operating on alpha times fee is alpha operating on TV that's the property of linearity and the property of continuity can be phrased in different ways but the most direct way of writing it now have a comment about this in just a second is that if TN converges to fee a sequence of test functions converging to a test function fee that implies that the measurements of fee n also converge to the measurement of the final test function so really there's nothing wrong with trying to impart sort of a physical interpretation but I don't want to try to push that too far but again you say that you think of fee is some sort of something to be measured and T is the thing that does the measuring now I want to make a comment about the mathematical the one one quick comment about the mathematical foundation for this the hard work in all of this is this statement right here fee n converges to fee right the hard mathematical work is to define TN converging to fee what that means because these space of functions although they each individual function has particularly nice properties infinitely differentiable all the derivatives decay and so on and so on if you want to talk about a sequence of functions converging then you're if the more you insist on your the more properties you insist a function having the harder it is to control the sort of convergence you know if the function is infinitely differentiable then you want all the derivatives also to converge if the function is rapidly decreasing you want all the derivatives to be rapidly decreasing and so on so it's hard to write conditions down that are going to control this that's what's difficult to do and if you open a math this is a warning if you open a math book that says something like theory of distributions or theory of generalized functions it's going to be like the first hundred pages or so are going to be devoted to analyzing what it means to say that fee n converges the fee and you look at that you say why would anybody learn this subject and I wouldn't blame you for saying that all right it's complicated there's no she weighs about if you're going to give a very precise treatment of it then you really have to do that we don't have to do that all right I'm going to take this sort of again as intuitive as you can make that imagine controlling the functions all the derivatives and so you can think in terms of the graphs the graphs of fee n converts to the graph of fee and everything is controlled is nicely as you want so suffice it to say it can be defined precisely but that's not for us to do all right but I did want to point that out the other bit of terminology here for those of you who have seen this for those of you who studied a certain amount of extra math and particularly the field that's usually called function analysis is that one says that the distributions are the set of distributions is the dual space the set of test functions that's another way of phrasing this so the distributions the particular class of distributions that you're considering are the doer is the dual space is our dual space the space of test functions bring that up the cocktail party some night watch the people scatter all right okay now we had a couple of examples last time I want to remind you of that the first example we had was Delta defined as a distribution and it's the simplest distribution this mysterious Delta function that's supposed to capture this property of concentration at a point emerges as simply evaluation and so examples Delta alright so Delta as a distribution is defined simply by Delta operating on fee is V of 0 all right this replaces now that's a pairing once again in all this you're what you're going to hear me say a lot today is I'm going to define a distribution what does that mean that means you give me a test function I have to tell you how to operate on it all right so you give me a test function fee to define a distribution I have to tell you how that distribution operates on fee so I say how does Delta operate on feet Delta operating on fee is just fee of 0 nothing else period all right this makes precise this dispenses with all of that about the classical definitely classical definition of Delta is e is 0 everywhere except at one point where it's infinity and the integral is one and so on and so on all that is gone all that is swept aside by this definition now you know it took a while to get to that definition okay but once it's there that's what you can work with okay likewise there's the shifted Delta I call it a Delta sub a and that's just defined by so again to define a distribution you give me a test function I have to tell you how its defined I have to tell you it operates and I say Delta a operating on fee is fee of a period period okay now I would be remiss if I didn't indicate there's no there's a graphical way of picturing these Delta's all right so the graphical picture and we'll use this to you usually indicate the Delta function by a spike at the origin Delta all right and since it's supposed to have infinite height I can't exactly draw an infinite spike so I just put a little arrow there everybody writes it like this all right and likewise Delta sub a is usually indicated by a spike at at a okay now something else all right since we've been so careful about this and since I made such a big deal out of the fact that it's all airtight and we can dispense with integrals and things like that I am NOT going to blame you and I'm not going to draw I'm not going to try to drive out of your system expressions like this integral from minus infinity to infinity of Delta of X V of X DX is equal to C of 0 if you want to write that ok I'll even write it from time to time it's ok nothing bad will happen to you all right among friends it's perfectly ok and a corresponding thing with it with a shifted Delta function it's alright all I want you to be aware of is that one can this is just really a mnemonic in some sense this is something some way more complicated way of writing that alright but in some formulas and in some contexts it's helpful to write that and ok do not be afraid now that you have the hidden knowledge all right what's really going on because this is what's really going on the second example we had of distribution was really an example of a whole class of distributions and that was my way of assuring us that nothing has been lost all right now these things are sometimes called as a generalized function so if they're generalized functions they don't include the original functions the good old functions and they do and is the second example I say a class of examples are really distributions induced by this is probably the best way of saying it induced by functions all right so if f of X is a function such that I can integrate it against the test functions such that this integral minus infinity to infinity say f of X fee of X DX makes sense the integral converges all right and that will be the case for many functions because fee is such a nice function fee may be such a nice function that even if F is a bad function the product is going to be such that the integral converges all right if that makes sense then that's a number after all I integrate f of X forces against fee of X and that's linear because the sum of the integral is the interval the sum and so on and it's also continuous that's a little harder to show and so that defines F have to find a pairing of f and fee continuous again I'm not going to show that a pairing of F and fie in that sense F defines or induces a distribution you give me a test function I have to tell you how the distribution operates on it how does it operate on it it operates by that pairing and so for shorthand I write F paired with fee when I'm really thinking here is the distribution induced by F but that's too many words so f of X times G of X DX okay so a lot of functions which are not as nice as the test functions are included in the set of distributions instead of operations things that operate on test functions so for example it may be the case that one pairs with fee so this includes many functions many functions induced distributions so can be regarded as distribution so eg it may be the case that integral of 1 minus infinity to infinity V of X DX which is the integral from minus infinity to infinity of 1 times V of X DX that may make sense it should make sense you don't want a function which is not integral for goodness sakes and so in that sense one induces a distribution because I can allow one to operate on fee by means of this inner integral okay and likewise sines and cosines are not themselves integral but if I multiply them by a function which is really dropping down which is decreasing at plus or minus infinity then the integral of sine or cosine against such a function will make sense and in that sense sine and cosine can be considered as distributions that is they operate them so let me do the just for short the complex exponential often works so e to the 2 pi ax is might be okay provided the interval from minus infinity to infinity e to the 2 pi ax fee of X DX makes sense which from any functions it will okay it will even though that even though into the two pi ax itself is not interval if I multiply by a function which is decreasing the product will be integral say alright so that induces a distribution how does it operate on the function it operates on the function by integration okay alright and if you look at this example and if you think of a Fourier transform coming in you wouldn't be too far off because now the big moment has arrived or I'm going to define the Fourier transform of a distribution so we've talked about distributions and how you the definition of distributions in a couple of examples of distributions I want to do something with them and this is so cool how this works I think so again just enjoy the ride alright and watch watch how these derivations go it's almost effortless almost effortless the way the way this all works I mean it's just so it's better than sex not that I would know I was a math major but so I want to define a Fourier transform of a distribution so now we're going to look the test functions we're going to look at now are the Schwarz functions the rapidly decreasing functions so let's take the test functions to be s the rapidly decreasing functions alright now let me remind you of why they are so good for Fourier transforms alright why they are so good for Fourier transforms is if V is an S and so is its Fourier transform all right and for that matter I guess I don't I don't think I said this but for that matter the inverse Fourier transform is also an S ok inverse Fourier transform differs from the Fourier transform only by a minus sign so if one isn't there it's easy to see the other one is so that's one property and the second is that Fourier inversion works all right it is in that sense those sense let me write the second one now that's the first property is that is that the functions rapidly decreasing so this Fourier transform second is that the inverse Fourier transform of the free transform physique one of the fee and likewise the Fourier transform the inverse Fourier transform physically those are the properties that make fee the right class the best class of the classical functions to use when you want to talk about the freight transfer the classical Fourier transform is defined by the interval all right so these are the functions we're going to use the corresponding class of distributions is sometimes called the class of tempered distributions of distributions are called the tempered distributions all right and what I want to do now is I want to show you how to define the Fourier transform of a tempered distribution to be another tempered distribution okay if T is a tempered distribution I want to define its Fourier transform and this will be another tempered distribution okay now again you will hear me say over and over again today I wanted to find a distribution what does that mean that means you give me a test function I have to tell you how it operates so I have to tell you how to define the Fourier transform something I'm going to call the Fourier transform of T operating on feet or paired with feet so I have to define before I transform of T operating on feet where fee is a test function rapidly decreasing function what shall I do it how shall I do it all right there is a guide in this there is a guide in all things all questions of this type you want to know how you should define an operation on the distribution you ask yourself you start off by asking yourself what would I do if I were in a really good situation here like if F of T were itself a rapidly decreasing function say or some other good function and this pairing was by integration alright so what if this is another example of suppose the problem is solved and see what has to happen alright so what if the pairing is by integration it may not be in general okay I mean Delta doesn't arise by integration but suppose it does arise by integration what would the formula where would I be led where would I be led so suppose that everything works out as smoothly as possible what are the consequences of that all right so let's say that before a transform paired with fee is the integral from minus infinity to infinity of a Fourier transform of T of X V of X DX alright so again this is this would not be necessarily the case in general because a distribution is not always given by integration but suppose it is and what happened what has to happen right so suppose then again I'm supposing that everything nice is happening so the Fourier transform is given by an integral it's integral from minus infinity to infinity e to used X here so let me write this minus 2 pi X Y T of Y dy then times f of X DX supposing that I can do this right supposing that everything is nice enough that I can do that all right then all right you know you buy the premise you have to buy the gag you have to follow your pencil all the way through and see where it leads you so then I can combine all that let me write this as one big integral minus infinity to infinity minus infinity to infinity then I'm going to split it up again so e to the minus 2 pi I XY T of Y f of X DX dy or I guess I wrote a dy DX dy DX doesn't matter because I'm now going to split everything apart again dy DX all right so now I'm going to swap the order of integration and I'm going to put the F did I call it a for fee sorry this is see if this is this is fee my apologies fee and this is fee ok so now I'm going to I'm going to swap the order of integration I'm going to put the fee with a complex exponential and leave the T alone alone and write this as the integral from minus infinity to infinity therefore minus infinity infinity e to the minus 2 pi X Y fee of X DX and then the result is integrated against T y dy now what is inside the integral everything is as nice as it could possibly be he said bouncing up and down so this is the integral this is the Fourier transform of fee evaluated at Y because I'm integrating e to the minus 2 pi X Y fee of X DX so this is the integral from minus infinity to infinity of the Fourier transform of fee at Y times T of Y dy ok where do we start where do we finish we started with supposing everything was as nice as it could be and the pairing was given by integration we had this expression the Fran transform of T paired with fee is the integral from minus infinity to infinity of T of Y before a transform of fee y dy and you have to look at this and look at this backwards in the sense that what appears here is the pairing of T and the Fourier transform with of fee all right if everything is as nice as possible and the pairing is given by integration then what has been what has occurred by our following our trusty pencil is that this is the pairing of tea with a Fourier transform of feet ok so now you say to yourself if everything were as nice as possible how would I pay our f of T with fee what would the Fourier transform of T be paired with fee would be the same as if t were paired with the Fourier transform feet okay all right now this right hand side is going to make sense regardless of what I did before I got to it all right why because tea is a tempered distribution tea operates on Schwartz functions he operates on rapidly decreasing function if V is a rapidly decreasing function so is its Fourier transform it's the property of that's the property of schwartz functions that's the property of rapidly decreasing functions so it makes sense for T to operate on this alright this left-hand side may not have initially made sense but there's a right-hand side will make sense right now bless your souls what is the thing to do thing to do is to turn the solution of the problem into a definition all right turn this into a definition define the Fourier transform so again I'm till I'm telling you so for you for T for a given distribution T tempered distribution T so T is a tempered distribution I wanted to find the Fourier transform I wanted to find a distribution you give me a test function I'll have to tell you how the distribution operates on that test function I want to define the Fourier transform of T by how it operates on a test function Fourier transform of T operating on fee is by definition T operating on the Fourier transform of feet on the right hand side this is the classical Fourier transform of fee given by an integral because Fiza nice the function is it could be this is a new F so to speak this is this is a definition of the Fourier transform of the tempered distribution T how do i define a distribution you give me a test function I have to tell you how it operates it operates by T operating on the freight transform on the right hand side makes sense because if V is a rapidly decreasing function then so is Fe I could smoke a quiet cigarette here but I don't smoke it's so cool now look you might say what a cheat I mean what a cheat I mean you're telling me after all this that the Fourier transform T operating fees T operating before H right that's what I'm telling you and that's because it's compelled to tell you that alright it has to work out that way if everything else is going to be consistent all right now how should we define the inverse Fourier transform we define the Fourier transform how should we define the inverse Fourier transform of a distribution well the inverse Fourier transform of T operating on fee must be nothing other than T operating on the inverse Fourier transform of feet okay what else is it going to be and now let's prove for the inversion for a inversion would say the for inverse Fourier transform the Fourier transform of T is equal to T for any distribution and also before a transform the inverse Fourier transform of T is equal T all right it's the most important subject is the most important theorem in the field all right if you can't invert the transform it's not going to do any good practically so this is the most important of theorem and it's trivial it's an absolute triviality why because everything has been carefully defined it is a triviality because all the terms in this expression how to operate with them have been precisely defined what is the inverse Fourier transform of the Fourier transform T operating on fie the inverse Fourier transform of something is equal to the something Fourier transform applied to the inverse Fourier transform of feet what is the Fourier transform of T operating on this test function which is legit because the inverse Fourier transform is again a rapidly decreasing function it is T operating on the inverse Fourier transform which is me on the issue it is T operating the Fourier transform of the inverse Fourier transform of feet Fourier transform operating on something is the something operating on affray transform its T operating the Fourier transform of the inverse Fourier transform of feet but in the space of functions which are as nice as possible for Fourier transforms classical Fourier transforms for a inversion works so this is T operating on feet where do we start where did we finish we have found that the inverse Fourier transform of the Fourier transform T operating on V is the same thing as T operating on fee for every test function fee and therefore the inverse Fourier transform the Fourier transform of T must be T period the deeper the most important theorem the subject emerges Apsley effortlessly because all the terms were properly defined the calculation just took care of itself all right the effort to try to prove this in the classical case is just murder aren't is just murder you can prove it in the case of Schwarz functions you can prove it in the case of rapidly decreasing functions and the proof actually is not it not in all detail but in most the details is given in the notes all right and there it goes pretty smoothly because the functions are as nice as possible but if you try to start to prove to prove for a inversion when the functions are not quite as nice as possible you run into all sorts of mathematical snares all right but if you broaden your perspective and that's what it that's what's really required here if you broaden your perspective on this and let this let these new ideas in then it becomes just a piece of cake it becomes a deduction immediately from the definition all right now let's have a Fourier transform hit parade let's calculate some Fourier transforms because one of things I said to you and I want to show you now in a way I hope you'll believe is that you can calculate with this definition it's not just for proving theorems it's not just for making things rigorous it's also provides you a way of calculating in a way that you can have greater confidence in your answers let's find the Fourier transform of Delta all right the mysterious Delta function now if it's what I mean trying to in trying to find the Fourier transform that I mean my god you know what could be harder in the classical case but in that when you give the set up in terms of distributions it is so simple watch to define the Freyja transform of Delta by definition the Fourier transform of Delta operating on on a function fee is Delta operating on the Fourier transform of fee that's the definition of the Fourier transform okay you give me a test function I have to tell you how it operates on the test function the Fourier transform Delta operating on fee is by definition Delta operating the for a transfer fee but now how is Delta operate on a test function Delta operates on the test function by evaluating that's the free transform of fee evaluated at zero okay but now what is the Fourier transform of fee at zero now I write down the classical definition of the Fourier transform it evaluated at zero I can do that because fee is just as nice a function is it can be this is by definition the integral from minus infinity to infinity of e to the minus 2 pi I 0 times X fee of X DX in other words is the integral from minus infinity to infinity of 1 e to the 0 1 times V of X DX and now you have to look at this and have to look at it with new eyes the integral from minus infinity to infinity of 1 times fee X is the function the Classen function one paired with fee if I regard the constant function one is inducing a distribution where do we start where do we finish we found that the Fourier transform of Delta prepared with fee is the same thing as 1 paired with fee and that is true for any test function fee and therefore the Fourier transform of Delta is equal to 1 the simplest of all distribution this has the simplest of all Fourier transforms the constant function one before a transform of Delta is one air tight air tight alright now you may very well have seen this in other classes you may very well have seen other sort of tortured derivations of this property that the Fourier transform of Delta is 1 but this is the right derivation of this property ok nothing is in question here absolutely nothing now you see how easily I can compute that all right I've got a certain amount of practice it's true but none of these steps is hard and it's worth your effort to sort of go through and say these things to yourself out loud say them out loud all right because it works so nicely and so easily by the way this is another example of this matter of fact this may be the extreme example of the sort of dual relationship between concentrated in one domain is spread out in the other domain all right Delta is infinitely concentrated all right sort of by definition that's what Delta is supposed to be Delta is supposed to be the function Delta supposed to be the limit of functions that are concentrating at a point it's Fourier transform one is uniformly spread out in that nice that's the other thing you have to sort of get say to yourself every time you have one I mean I know it's hard because there's so many little bits and pieces like this but that's one of the things you start to get used to when you work in the subject for a while when you when you when you use it is you start to have these little checks you know these little little interpretations that you carry with you from old situations to new situations the old situation was in the stretch theorem concentration the time domain means spreading out in the frequency domain and vice versa alright so here concentration the time domain is spread out in the frequency domain in the in the most uniform in the most extreme case let's do some more what is the Fourier transform of Delta sub a the shifted Delta function all right what about the ship to Delta function Delta sub a all right what is the freight transfer the Fourier transform of Delta sub a I have to tell you how it operates on a test function the Fourier transform of Delta sub a is by definition Delta sub a operating the Fourier transform of feet but what is Delta sub a operating on anything it evaluates the thing at a that is this is the Fourier transform of fee at a alright what is the Fourier transform of feet at a it is the integral from minus infinity to infinity e to the minus 2 pi ax fee of X DX and now again you have to look at this with new eyes and you have to say to yourself this is the pairing of the complex exponential with fee this is the pairing of e to the minus 2 pi ax with fee where do we start where do we finish we have the Fourier transform of Delta sub a paired with fee is equal to e to the minus 2 pi ax paired with fee and therefore that holds for every test function fee that identifies the Fourier transform of Delta sub a as e to the minus 2 pi I a X where the right-hand side is understood as a distribution okay it's the distribution which operates on a test function by this pairing you may have seen this you may have seen some tortured derivation of this too but this is the right derivation of it and while we're at it why don't we find the Fourier transform of the exponential complex exponential hey hey hey let's find the Fourier transform where is it here yes let's find the Fourier transform of e to the 2 pi X how do I do that where this is understood as a distribution this would not exist in the classical case right I mean you can't find the Fourier transform of this by computing the integral but we can by pairing before I transform you the two pi ax paired with fee is equal to by definition e to the 2 pi ax pair - the Fourier transform fee that is the integral from minus infinity to infinity of this is a function pairs with Fourier transform e to the 2 pi ax period of the Fourier transform fee X DX now look at that and what do you see what do you see what do you see you see the inverse Fourier transform of e classical Fourier transform you are computing there the inverse Fourier transform evaluated at a this is fee of a because this is the classical inverse Fourier transform of the Fourier transform at a so it is V of a but fee of a is the Delta function at a paired with feet if you use your new eyes alright so where do we start where do we finish we started with that we finish with that the Fourier transform of e to the 2 pi ax paired with fee is the same thing as Delta sub a paired with fee that identifies the Fourier transform e to the 2 pi ax as Delta sub a air-tight no question take the particular case actually when a is equal to 0 take the case when a is equal to 0 if a is equal to zero I have a Fourier transform of 1 e to the 2 pi 0 and what do I get the Fourier transform of 1 is Delta sub-zero so the Fourier transform of Delta is 1 the Fourier transform 1 is Delta no doubt about it no doubt about it and again this is a nice illustration of spread out in the time domain concentrated in the frequency domain uniformly spread out in time infinitely concentrated in frequency Delta concentrated infinitely concentrated all right it works it works great couple more a couple more how about do it up here sines and cosines sines and cosines do not have classical Fourier transforms but they have generalized Fourier transforms because sines and cosines makes sense as distributions all right for instance cosine of two pi ax I'll stick with a sort of scaling by a alright that's easy to do because that's one-half e to the 2 pi ax plus e to the minus 2 pi ax then I take the Fourier transform of each part each piece so the Fourier transform of cosine of 2 pi ax is 1/2 the Fourier transform of this which is Delta sub a plus the Fourier transform of this which is Delta minus a or place a by minus a in this expression so it's Delta sub minus a ok so simple so simple really there's a graphical way of representing that the graphical picture usually goes like this you write up by two spikes one at a and one at minus a near zero so this is usually this is this is people usually denote the Fourier transform graphically by this and that's fine you know that's fine you can write it like that the picture is that how about the sign the sign function is just as easy because the sign function can also be expressed in terms of complex Exponential's the imaginary part of the complex exponential so the sine of 2 pi X 2 pi ax is 2 pi a X is 1 over 2i times e to the 2 pi ax minus e to the minus 2 pi I a X and so it's Fourier transform which I guess I'll do over here is the difference of two Delta functions times this complex number 1 over 2i Fourier transform of sine of two pi ax is 1 over 2i times Delta sub a minus Delta sub - a nothing to it nothing to it alright once again I said a little while ago that the problem with the classical Fourier transform is that it didn't make sense on the functions that you really wanted to make sense on like the functions that society needs trig functions constant functions and so on alright but now if you open your mind and open your eyes all those things work all those things work and it's and it's effortless it's effortless now it takes some effort before it becomes this effortless yeah we have this more general framework is that also how you define a Fourier transform of classical functions that you can do before like the rect function all right so so that's a good question the question is what happens to wall to wall to all that we did before now honestly to be to be the most honest and rigorous about this you should always consider tempered distributions so that is to say if you want to take the Fourier transform of the rect function all right you can't there's two ways of doing you can consider it classically that's fine there's nothing wrong with that or you can say the rect function induces a distribution right and it's Fourier transform is the sinc function also understood as a distribution now that's a little that's going a little too far and some in some sense in that you know the when the functions are nice enough you do what a function induces a distribution you tend to blur the distinction between the function and its dist and the distribution that it induces alright but the most proper way of saying it is you only consider the Fourier transform for tempered distributions that's that's where it's properly defined and so classical functions where we computed the Fourier transform before should be understood as distributions and everything should be in that framework all right but then you have to say to yourself look we're not kids here we have to take a reasonable approach of this and you don't want to abandon what you have done in the past all right so it's one of these things where you have to keep a little tension a little cognitive dissonance right in your head between the Fourier transform on tempered distributions and the Fourier transform on a function when it does make sense alright so and I am and I'm going to make that blur I'm going to blur line all right that is I'm going to write the Fourier transform the rect function is the sinc function without worrying about it all right without saying really it has to be distributions as to be understood in the sense of distributions and so on so for example when you write now a matter of fact and I say this in the book you can finally justify completely that the Fourier transform of the sinc function is the rect function or the inverse Fourier transform of the sinc function is there is a rect function same thing all right you can do two ality you can do all those things the proper way of understanding this is in the sense of distributions because the Fourier transform the sinc function exists sinc exists as a district induces a distribution it's affine function as far as integrating against the Schwartz function goes alright so the Fourier transform the sinc function is a distribution this has to be understood in the sense of equality of distributions not in not in terms of sort of point wise functions but that gets a little extreme alright that gets a little extreme you know even even religion has its limits all right now I'll say one more thing one more thing like I said I just wanted you to enjoy the ride a little bit there's a little bit more of the ride to go tomorrow next time I'm going to talk about derivatives of distributions how every do every distribution is differentiable removing that flaw from the classical theory of calculus we're not every function is differentiable every every distribution is differentiable and we're going to get some formulas out of that um you know why do we go why do we go through this you know again you have probably seen these things derived in some tortured way in previous classes all right well like I say things have moved on this is the modern way of viewing Fourier transforms and how they work and you should as part of your education want to know what the modern view point is you could build a radio out of vacuum tubes but we don't teach that anymore alright we don't teach a vacuum tube technology all right you should want to know even if only to be even only to have sort of coins that it's possible to do you should want to know how the modern view of this is and furthermore as I say it's not just that it's a good point of view you can actually compute with it with confidence I derived effortlessly these formulas like the Fourier transform one is Delta the Fourier transform of Delta is one so the derivation is those in the classical case are really quite involved and not at all I think to my mind convincing all right they work they get you the correct answer but you know the cost of trying to of torturing things to do that is quite high all right I see people wanting to get in we're going to take a break I know the others question in the back you can ask me when we're on way out okay all right

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Keywords: Electrical, engineering, computers, math, physics, geometry, algebra, calculus, technology, functions, linear, operations, sin, cosin, Fourier, transformations, series, distribution, test, signals, rapidly, decreasing, generalized

Id: s99rd_tIozY

Length: 49min 25sec (2965 seconds)

Published: Thu Jul 03 2008