Intro to Fourier transforms: how to calculate them

hi again everyone Chris Tisdell here again in this presentation I'm going to start a series of videos on Fourier transforms now the nature of this particular presentation is extremely elementary we are going to define what a fourier transform is and the inverse Fourier transform and we're going to calculate a few basic transforms now interesting theorems associated with the Fourier transform and it's applications to partial differential equations will be seen in other videos so the nature of this particular presentation is extremely elementary now the big motivation for studying for it transforms is like many transform techniques such as Laplace transforms the motivation is to try to solve differential equations both ordinary and partial now the ideas that we're going to look at in this section are going to complement the transform methods that you might have seen with the plastron s-- forms now if you know a little bit about Laplace transforms the transition to fourier transforms is not a huge jump now Fourier transforms apply when limits are up source fully transforms appear when limits are applied to Fourier series and in particular one of the limitations of Fourier series is that they apply to functions that are periodic because they involve a sum of periodic functions cosine and sine so this naturally raises the question what about non periodic functions and Fourier transforms can accommodate these kinds of non periodic functions where Fourier series can't handle them so this is something that Fourier transforms can do that Fourier series cannot do now I'm not going to show you the derivation or the suggestion of how Fourier transforms arise from Fourier series I'll leave that for another video but essentially what's happening is you're letting the period of the function approach infinity okay so what is a Fourier transform okay well suppose I've got a function of say one variable f of X and that function is piecewise continuous and absolutely integrable on the whole real line so by absolutely integrable I mean the following this improper integral converges then we define the Fourier transform of F by this integral transform here okay now there's just like with Laplace transforms there's two ways of really denoting the the transform - through this curly bigair for cap calligraphic F the other is through this F hat now this particular notation emphasizes the dependency on w if you look over here you're done with variables X you've got I remember I squared equals negative one and you'll be left with some function of W here okay so this notation emphasizes our original function this notation emphasizes that the transform really is a function of W and there is a partner to our for it transform defined through the following well we have this curly F inverse and if you apply this curly F inverse to to this you should recover the original f of X okay so this is also an integral type expression now some books have different coefficients here then it's not a big deal the important thing is that the product of these coefficients should equal 1 on 2 pi ok so here I've got the square roots in there just because there's a little bit of symmetry associated with with each definition then ok so it's not a big deal okay so let's look at an example here's one here we asked to let a be a a a positive constant compute the Fourier transform of the following function okay well if I'm going to draw a little graph here of the function it's going to look something like this okay satisfies these these conditions here let's see if we can just apply the definition and compute the integral and and the the Fourier transform okay so by definition the Fourier transform of a function is just this okay now now how FX is defined in a piecewise way on this interval it's defined like that and on this interval it's defined like that so let's break up the integral into two bits and we'll calculate both of those integrals okay so oops okay negative 52 0 FX on that is the following so it's going to be e to the ax so I can combine it with the following so I'll get that and similarly let's integrate from 0 off to positive infinity and it's going to be the following which I can team up with my exponent there and it's going to be something like this okay so now it's just a matter of knocking over these improper integrals and then we'll be done okay so imagine a Omega and is it's essentially a constants because we're integrating with respect to X so we'll get this and same with this one okay so now it's just a matter of plugging these limits in or these values in and seeing what we have okay so when x equals zero we're going to get the following now as X approaches negative infinity what's going to happen well this will go to zero okay essentially what's happening there is the e to the a X because a is positive and X is large and negative is sort of dragging everything in this term to zero okay and if you want to verify that you can actually write say e to the negative I I W X as cosine W X minus I sine W X that's a bounded function okay and so if i multiply by e to the ax and X goes to negative infinity that product is going to go to zero okay same over here the as X approaches positive infinity this is going to approach zero and when x equals zero the top is going to be one and I'm going to get something like this okay so now I can just sort of clean this up a bit okay so I'm going to get the following plus okay and now can even go a bit further and get rid of the the eyes just by cross-multiplying okay so that times that Plus that times that all over that times that okay so we make some cancellation out here when you expand this bracket remembering I squared equals negative one will come up with the following okay so the Fourier transform of this function is this now what does it what does it mean with respect to the inverse transform well the inverse transform of this is what we started with this function okay okay so that's a simple example of calculating the Fourier transform of this f of X okay let's try a slightly more challenging problem now here we asked to compute the Fourier transform for this particular function and remember this is our definition of the Fourier transform so let's work our way through this one and this will be a little bit more work okay and the reason is we've got an x squared here we did an example before where there was basically just an X up here well the x squared makes it a little bit more challenging okay so let's use our definition so by definition the Laplace transform or sorry the the Fourier transform of a given function is the following integral representation okay and so let's replace f of X with this and then we'll try to knock over the B integral okay so this is also an exponential function so what I can do is team that up with the existing exponential function and I'll get the following okay now how do I evaluate this integral work it's difficult because of the x-squared up there okay but what I'm going to do is complete the square in X and then that's actually going to simplify things quite quite a lot okay so so I'm going to complete the square in X okay so essentially I've got something like the following in the inner brackets here okay managing the AOA so take half the coefficient of x square it add it and take it away again out these returned are perfect square so I can factor those and over here expanding this say you'll get I squared equals negative one so you'll get something like the following okay alright let's put that back into here and see if that really does simplify our situation okay so II to negative of that well the X remember integrating with respect to X here the exes are here that's got nothing to do with X so I can break it up into something like the following sorry it should be a squared down okay all right don't forget that should be a squared down okay so times okay so the the a will cancel with that a and I'll get the following now this has nothing to do with X so I can push that sorry flip minus one so I can push that out the front okay now the question is how do I evaluate this now it's an improper integral and it's e to the negative I guess of something squared essentially now this almost looks like an error function which is a special function used in applied mathematics okay so at this stage we note that we know the following I can I've proved this result in other videos in fact I've proved the result from 0 to positive infinity is this on - okay it's an even integrand so to go from negative infinity to infinity you just multiply the result by 2 so that's where the root Pi comes from ok so what I'm going to do is make a substitution up here ok I want I want that to become a negative Z squared ok so I'm going to let say Z equal to square root of everything up here okay so if that's the case then D Z is going to be root a DX the limits of integration are going to stay the same so our transform is going to be something like the following uh this might change uh-huh okay so I'm going to eat of the negative Z squared times DZ over root a okay so that can come at the front and then I'm left pretty much left with this okay now the now if I move that at the front this from here is just root pi so the root PI's will cancel and I get the following okay okay so out transform then is the following okay that was a little bit more work than the previous question now a good question here is why why is this result important okay why is this an important transform to spend some time learning how to work out well first of all what you know one answer to that is that you know we want to get familiar with with a transform technique okay but why is this particular function important well this particular function is important because you you'll see it when we're solving partial differential equations and in particular when we studying diffusion or heat problems involving PDS okay we might see the transform or to the inverse transform it essentially the solution method to heat diffusion problems from partial differential equations okay so that is the that is the significance of this particular example and we'll rely on that a bit later when we do solve some of those problems so that's my presentation I hope you found the introduction nice and gentle when you confident with the idea of a Fourier transform or just the computation of it of course this result implies that the inverse transform of this is is this okay now please join me for more videos in the future where I'm going to build on this basic presentation of Fourier transforms I'll talk about more theory interesting theorems involving transforms and of course their application to partial differential equations please join me for those presentations

Channel: Dr Chris Tisdell

Views: 189,388

Keywords: Fourier Transform (Idea), math, pde, partial differential equations, Chris Tisdell, example, education

Id: WcNPUXfxCXA

Length: 22min 7sec (1327 seconds)

Published: Fri Sep 27 2013