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So the aim for this lecture is say we forget about these frequencies.

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We just have this plot here just as data, and we want to figure out what are the three frequencies

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that are superimposed here.

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And for this, we will use these so-called for your transform.

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I will not explain to you, why did Fourier transform will give us the characteristic frequencies?

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We just assume that this is correct is is something one can derive in mathematics.

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And here we want to just implement this and test how it works and see that it really works.

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So the equation here means that we have some intensity, which I call Whitehill, but it doesn't really

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matter.

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You could give it any name you want, and it's a function that now doesn't depend on the time, as previously

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we had why of Ti, but now it depends on the frequency omega.

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And it is calculated by calculating such an integral over the time.

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And actually, we integrate here from minus infinity to plus infinity, which of course we cannot really

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do here.

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So we must be clever about these values and find some tricks.

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And then we integrate over our system or a signal, which will be this one here.

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Of T.

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And then here's two factor that I mentioned it's an exponential function with an imaginary argument.

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So this will be the imaginary unit I, which is the square root of minus one.

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And then we have here the omega, which is the argument of our function and the time.

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So it works like this.

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You basically take a value for Omega, for example, zero point five and then you put it in here.

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And then from this, you calculate this integral, then you go to the next value of 0.6.

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Calculate again.

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And so you will get a function y tilde of omega, which we will let them have a look at later.

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And I can tell you already that this function omega tilde is no sorry y told off omega will give you

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a very pronounced peaks at the frequencies that we can see here.

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So let's go ahead and implement this.

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We need first of all, Omega is equal to, as an example, 0.3 and then into Grant, which will be this

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one here is basically given S and P dot array.

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And then.

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T list

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like this.

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And then one over and p dot square root two times and p dot pie than we have.

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So I basically put this into the integral doesn't matter if it's outside or inside.

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I put it inside now and then we multiply by the wine list and then we have to value and pedaled.

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Exp.

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The eye is written in Python as 1g times omega times t list.

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And then to calculate the value of tilde of Omega, we just write print integral Trump P is sidle of

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integrins.

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So basically, this means take this array that we have here where our x axis, so which is the time

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is still the same and our Y value is changed from y to this whole integral tier.

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So one of a square root of two pi times point of T times e to the power of Iomega T and then use the

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integral trapezoidal method to this and this will give us just a number.

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So you see, it gives us actually a complex number.

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So a real part and an imaginary part.

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So we probably need something like this absolute value of this.

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And then typically you just square it.

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This is just like typical practice.

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OK.

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And so the value would be done.

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Four hundred twenty.

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That's just a coincidence.

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OK, so we have now calculated here actually why till the absolute value squared, which will be what

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we will plot in the following.

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So this is just a value for one particular omega 0.3.

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And now we want to do it, of course, for the whole range of Omega that we are interested in.

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So we should probably start at zero because negative frequencies are not so important for us.

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Here we take only positive frequencies and then we go to a value where we assume after this we don't

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have any peaks left.

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If we look at this spectrum here, we would say, OK, what is here the fastest oscillation, which

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is the smallest period?

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So this will then have the smallest frequency.

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So probably something on this length scale here.

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What timescale better to see?

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And we know already since we know the solution that there is no frequency larger than this one.

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So probably it's OK to just look in the range of zero to five and just to be on the safe side, we will

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look in the range of zero to 10.

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So I will right now 50, which will be the data that we will plot later on, is own rate and the rate.

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And first of all, first lists for the first entry of the list will be Omega, which I call on Omega,

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and it doesn't really matter here.

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And then the next value will be basically what we have written down here, integral trapezoidal acting

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on basically this one.

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So let me copy this.

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So we have integral trapezoidal acting on ENPI array tea list, the square root.

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Whilst the exponential and here we just have to change this to on and now we must specify what it is

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omega means.

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So for this, we have to add here another bracket and like this and then we can write for omega in and

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p total in space, starting from zero going to 10.

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And then we must specify how many steps we want to have in our omega list.

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So I don't know.

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1001.

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All right.

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So this took maybe one second to calculate and now we can go ahead and plot.

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So we have on the x axis the frequency

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omega and on the y axis

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we have the intensity which will actually be absolute value of to the square.

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And then we plots peel t dot plots for your transform.

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This is our list that we have just calculated after all the values.

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Just the first component.

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So basically this list here

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and then absolute value of f t all the values come on one and the whole thing squared as before, just

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like this one.

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And now we run into a bit of a problem here.

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This is because it says complex warning, so it's not an arrow is just a warning, um, testing a complex

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value to a real function.

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This called the imaginary part, and I checked before.

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What's the problem here?

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The problem is actually this list here.

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So the list of Omega's, because the list of Omega, since we have here a complex number, it assumes

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that these may also be complex numbers.

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So this seems to be, for example, zero point zero two plus zero I so always it adds a zero I, which

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of course, doesn't mean anything, but it changes the data type to a complex number.

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And then if you want to plot something like this gives us a warning that the imaginary part is disregarded.

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So in our case, it's really what we want because there is no imaginary part, it's zero.

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But just to get rid of this warning, we could write Dot Real, but it's not at all necessary.

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So now we have our results.

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This is the intensity spectrum over the frequency omega.

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And you see we have some pronounced peaks here.

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One of them is out to the other one at 3.5 and the other one is here at a very low frequency.

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Hard to say.

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We would have to analyze the data, but since we know what it is, it is probably 0.3.

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So we have now high frequencies to 3.5 and the other one 0.3 we can figure out just in a second.

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We basically just rides f t and then all of the values basically from let's say we know that it's at

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zero point three.

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So let's look at this one.

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So this is the range of the points corresponding to Omega from 0.25 to 0.34.

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And now, of course, we have to look at the Y values and the absolute values squared.

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So maybe you could, of course, make this in a more elegant way, but just to do it quick and dirty,

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you could do it like this.

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Absolutely, you squirt.

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And now you must see here, OK, it increases, and then here we have the maximum and then it decreases

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again, and the maximum corresponds to six one two three four five six zero point three.

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So this is really the frequency that we see here, and this is corresponding to the frequency here.

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So you see really the proof that by using a free transform, we can calculate the specific or characteristic

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frequencies of a spectrum.

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And it means that our function y of T is superimposed by individual periodic functions that can be cosine

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functions that can be sine functions that can be cosine functions with a phase shift.

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All of this doesn't matter.

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We just are able to figure out the individual frequencies and later in the course, in a later section

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about eigenvalue problems, we will really use a Fourier transform to figure out the characteristic

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frequencies of a problem where we don't know what the characteristic frequencies.

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So this will be really exciting then, and it can really be a useful tool.