1 00:00:00,330 --> 00:00:08,130 So the reason why we used the eigenvalue methods to analyze the EIGEN frequencies was because we wanted 2 00:00:08,130 --> 00:00:14,550 to analyze the numerical result a bit more because it looked pretty difficult and we couldn't really 3 00:00:14,550 --> 00:00:15,870 learn much from it. 4 00:00:16,830 --> 00:00:22,920 But now, since we have the ion frequencies, it would be a good thing to figure out if we can really 5 00:00:22,920 --> 00:00:28,380 find these frequencies, which are given by just groups of these numbers here. 6 00:00:29,100 --> 00:00:35,750 If we can really find these in the numerical solution and this is what we will do next. 7 00:00:35,760 --> 00:00:39,960 So first of all, let me copy of this one here because we will need this. 8 00:00:41,280 --> 00:00:49,860 So like this, and the first thing that I will do is here is once again, the equation for the full 9 00:00:49,860 --> 00:00:56,250 year transform that we will use now, and we can use the four year transform to find the characteristic 10 00:00:56,250 --> 00:00:59,200 frequencies of such a spectrum here, which is harmonic. 11 00:00:59,670 --> 00:01:02,400 So sorry, not harmonic, but periodic. 12 00:01:02,850 --> 00:01:11,250 And we can basically split the non harmonic spectrum into several harmonic spectra, and it will be 13 00:01:11,250 --> 00:01:13,530 then a superposition of these contributions. 14 00:01:14,850 --> 00:01:19,560 So the first thing that we need is something that we have programmed much earlier. 15 00:01:20,130 --> 00:01:23,850 It was a method how to calculate integrals. 16 00:01:24,450 --> 00:01:31,920 And we did several different methods, and one of the better ones was the trapezoidal methods. 17 00:01:32,580 --> 00:01:34,980 So we have programmed this already in a previous lecture. 18 00:01:34,980 --> 00:01:39,660 Here, I will just type it in case you forgot it. 19 00:01:39,660 --> 00:01:47,310 You can also go back to the previous Notebook Python book and just copy the code that we have written 20 00:01:47,310 --> 00:01:48,060 that down there. 21 00:01:48,060 --> 00:01:49,140 So it's the same thing. 22 00:01:49,830 --> 00:02:01,470 So basically, we loop over all the points except for one, and then we add up the contributions which 23 00:02:01,470 --> 00:02:03,090 are given as follows. 24 00:02:03,660 --> 00:02:10,979 So a plus and then data the Y. 25 00:02:10,979 --> 00:02:11,760 Well, you. 26 00:02:12,390 --> 00:02:14,670 And then the next index. 27 00:02:18,080 --> 00:02:21,650 Adds to my value of the current index. 28 00:02:23,640 --> 00:02:29,970 Divide it by two, and then we have here the difference in the X values. 29 00:02:31,860 --> 00:02:38,570 So this would be day to zero five plus one and data. 30 00:02:40,480 --> 00:02:43,540 Zero calories, so think about it for a second. 31 00:02:43,900 --> 00:02:49,660 And if you really don't remember how it works, then you can go back to the video where we programmed 32 00:02:49,660 --> 00:02:52,240 this, it was in the integration part of the course. 33 00:02:53,530 --> 00:02:53,760 OK. 34 00:02:53,770 --> 00:02:56,050 And of course, in the end, we return this summer. 35 00:02:57,070 --> 00:03:05,800 OK, so this was how we can integrate a dataset where we have basically two lists a list for the X component 36 00:03:05,800 --> 00:03:07,660 and the list for the Y components. 37 00:03:08,650 --> 00:03:10,240 And we want to use these methods. 38 00:03:13,720 --> 00:03:19,480 Like this well, to use this method to calculate this integral here, which is if we transformed. 39 00:03:20,470 --> 00:03:27,220 But before we do this, what we can also do is we could integrate the individual fluctuations here. 40 00:03:28,520 --> 00:03:32,440 OK, but this you can do as a practice if you want. 41 00:03:32,920 --> 00:03:36,920 What I will do next is I will generate the data, right? 42 00:03:37,510 --> 00:03:46,480 So as I said, it's an array of X and Y values and it will be and he dot free. 43 00:03:48,730 --> 00:03:56,140 And now basically, we have to access the solutions. 44 00:03:56,140 --> 00:04:03,370 Here we have basically here we have the Tier T list for the home values, which will give the X component. 45 00:04:03,850 --> 00:04:09,400 And then we have these solutions, which will be the coordinates, which will be the Y solution. 46 00:04:10,360 --> 00:04:17,980 So basically, we will specify now a list where the X components will be just a time list. 47 00:04:18,550 --> 00:04:19,690 So that's pretty easy. 48 00:04:20,050 --> 00:04:26,400 We can just write and p dot a. 49 00:04:29,880 --> 00:04:34,350 Solution three o a seed dot t. 50 00:04:35,130 --> 00:04:37,170 And now we must try to come up. 51 00:04:37,570 --> 00:04:42,720 And now we must write the the function of which we integrate. 52 00:04:43,300 --> 00:04:48,570 And this would be one over square root of two pi times the solution. 53 00:04:48,660 --> 00:04:52,290 So basically this one and then this exponential factor here as well. 54 00:04:53,580 --> 00:05:02,760 So it will be, as I just said, one over and peed on the square root two times and peed five times 55 00:05:03,060 --> 00:05:04,320 the solution. 56 00:05:04,320 --> 00:05:07,260 So this would be solution on the score. 57 00:05:07,290 --> 00:05:10,230 Three O C Thought Y. 58 00:05:10,770 --> 00:05:16,070 And now we could, of course, program the individual components each one after another. 59 00:05:16,080 --> 00:05:19,180 But here I want to make basically an array inside of an array. 60 00:05:19,200 --> 00:05:20,540 This is why I wrote here also. 61 00:05:20,540 --> 00:05:20,880 Three. 62 00:05:21,450 --> 00:05:24,300 So we will later on loop over this index here. 63 00:05:25,110 --> 00:05:26,910 So we have now why? 64 00:05:27,570 --> 00:05:30,520 And then the IV component, which will be zero, one and two. 65 00:05:31,170 --> 00:05:40,920 And then we multiply by the exponential function and p dot p s p and then the imaginary units, which 66 00:05:40,920 --> 00:05:54,480 in Python is one g times o m, which will be the frequency and then times the solution. 67 00:05:56,460 --> 00:05:59,790 And basically, this will be time again, so I can just copy this one here. 68 00:06:00,990 --> 00:06:05,100 So you see this one is eye times, omega times t. 69 00:06:05,940 --> 00:06:16,200 So exactly what we wrote here and what is important to realize or to remember is that free to transform 70 00:06:16,680 --> 00:06:19,800 is a function depending on omega. 71 00:06:20,040 --> 00:06:24,660 So for each value omega, we have to calculate this integral here. 72 00:06:25,590 --> 00:06:30,900 So of course, we could now say, OK, we want to use Omega is equal to sun number, for example, I 73 00:06:30,900 --> 00:06:32,040 don't know 0.5. 74 00:06:32,430 --> 00:06:37,500 And then we calculate the value of the four you transformed for this particular omega. 75 00:06:38,130 --> 00:06:41,310 But actually, we want to have it as a whole array. 76 00:06:41,820 --> 00:06:48,510 So I will just write here integral trapezoidal. 77 00:06:50,960 --> 00:07:02,060 And then we have now our array or data set basically with the X values and the Y values over which we 78 00:07:02,060 --> 00:07:02,770 integrate. 79 00:07:02,780 --> 00:07:09,530 But I just said Omega is just a fixed value, so we must create a whole array where we basically have 80 00:07:09,530 --> 00:07:17,810 different values for omega so we can write like this omega comma and the value. 81 00:07:18,320 --> 00:07:28,490 And then we can basically just tried to close it here and write four and then we can write for Omega. 82 00:07:31,260 --> 00:07:34,650 In A.P. Dodds, Lance Bass. 83 00:07:35,070 --> 00:07:45,240 And now we must specify the range for Omega, and I will take values from zero to three and steps of 84 00:07:46,440 --> 00:07:48,360 size zero point zero one. 85 00:07:49,410 --> 00:07:51,990 So this will now be a loop over the Omega. 86 00:07:52,650 --> 00:07:55,810 But I think we are not finished yet. 87 00:07:55,850 --> 00:07:59,890 Yes, this is because we have to close. 88 00:07:59,910 --> 00:08:02,130 Now, of course, the brackets I will do this year. 89 00:08:05,650 --> 00:08:19,990 And opened the bracket, and then I want to loop over another index, which will be for I in range three 90 00:08:19,990 --> 00:08:25,220 because we want to loop over all three oscillators and here we have to index it. 91 00:08:25,240 --> 00:08:29,170 So we need basically another one of these brackets here. 92 00:08:31,540 --> 00:08:38,409 OK, so now we have F.T, which is the whole array of the Fuji transform, and we can, first of all, 93 00:08:38,409 --> 00:08:45,940 say, for example, F.T 051 F2 and then four three will give an error. 94 00:08:46,070 --> 00:08:51,840 Yeah, because this is from this loop here, this will just be the index for the oscillator. 95 00:08:52,420 --> 00:08:57,670 For example, if we want to have the free transform of the first oscillator, we wrote F.T Zero and 96 00:08:57,670 --> 00:09:08,770 then when we arrived one, when we write this, then we have all the the values of the four transform. 97 00:09:08,780 --> 00:09:12,880 So basically this one here y tilde of omega. 98 00:09:13,540 --> 00:09:17,950 And if we write zero, then we have all the Omegas. 99 00:09:19,060 --> 00:09:24,490 So you see here they are still stored in the complex data type, but actually it's just a real number 100 00:09:24,490 --> 00:09:27,280 because the imaginary part is always zero. 101 00:09:28,360 --> 00:09:32,920 So now we can go ahead and plot this, so I will write 102 00:09:35,410 --> 00:09:38,050 Potti Dot plots. 103 00:09:38,830 --> 00:09:42,850 And as I just showed you, we have to now write the omega lists. 104 00:09:42,860 --> 00:09:49,300 So this would be hefty zero comma, all comma zero. 105 00:09:49,780 --> 00:09:53,250 And then to get rid of the imaginary part, we can write dot real. 106 00:09:53,830 --> 00:09:57,190 And then we must use the actual value. 107 00:09:57,210 --> 00:09:59,080 So here we must change this to one. 108 00:09:59,560 --> 00:10:05,920 And typically, when you plot, if we transform to take the absolute value because it's a complex property 109 00:10:05,920 --> 00:10:07,350 and you square it. 110 00:10:09,610 --> 00:10:11,590 OK, now I will add the labels. 111 00:10:12,010 --> 00:10:16,870 Plus, he told X label is frequency. 112 00:10:20,340 --> 00:10:28,380 Well, it's actually frequency Omega, so this would be two pi times the frequency and then Kielty dots, 113 00:10:29,010 --> 00:10:31,260 why label is? 114 00:10:31,650 --> 00:10:38,220 You could see something like Inten City of Tafoya transforming. 115 00:10:41,110 --> 00:10:42,470 OK, so here we go. 116 00:10:42,490 --> 00:10:51,940 This is the fit, and now we could continue and rides peel to. 117 00:10:53,940 --> 00:10:54,820 Talk show. 118 00:10:57,180 --> 00:11:01,880 And now we can basically plot all of three free transports. 119 00:11:03,630 --> 00:11:08,600 So this would be done one two one two. 120 00:11:10,920 --> 00:11:18,210 And you see, these are now all the four you transforms with individual peaks at certain frequencies. 121 00:11:18,510 --> 00:11:25,950 So what's worth highlighting here is that these three peaks here correspond to the three frequencies 122 00:11:25,950 --> 00:11:30,150 that we have discovered before I have plotted them or I've written them down here. 123 00:11:30,630 --> 00:11:34,410 So this was what we have figured out eigenvalues eigenvectors. 124 00:11:34,860 --> 00:11:41,490 And when you calculate the numbers here, this gives 0.7, one point four and one point eight. 125 00:11:41,790 --> 00:11:45,520 These are exactly the values that we see here for the peaks. 126 00:11:46,200 --> 00:11:52,620 So indeed, if we calculate the four transform of our numerically solved equations, we get peaks at 127 00:11:52,620 --> 00:11:56,490 these unique frequencies, which are two agan frequencies of the system. 128 00:11:57,630 --> 00:12:03,900 And furthermore, we can also have a look at the eigenvectors because we have seen that the first and 129 00:12:03,900 --> 00:12:07,360 the third oscillator, they move in a pretty similar way. 130 00:12:07,380 --> 00:12:15,660 They often only change or have different signs here and here and here, and then the middle oscillators 131 00:12:15,660 --> 00:12:24,600 special because, for example, for the EIGEN mode with omega equal to one point four, the eigenvectors 132 00:12:24,600 --> 00:12:25,200 zero. 133 00:12:25,650 --> 00:12:33,240 And indeed, for the second oscillator, we do not see a peak in the frequency spectrum at one point 134 00:12:33,240 --> 00:12:33,600 four. 135 00:12:34,200 --> 00:12:40,590 So for this particular oscillator, this frequency does not exist, and we know that this is true because 136 00:12:40,590 --> 00:12:45,330 the eigenvectors has a zero for the component for this particular frequency. 137 00:12:46,080 --> 00:12:51,210 And also, we see that the spectrum for the first and the third oscillator are exactly the same. 138 00:12:51,990 --> 00:12:59,400 So our results from the eigenvalue problem are indeed contained in the numerical solution, as we have 139 00:12:59,400 --> 00:13:05,220 just analyzed by calculating the Fourier transform of this spectrum here. 140 00:13:06,600 --> 00:13:15,720 So what this means is that we can express the numerical solution by a superposition of three different 141 00:13:15,720 --> 00:13:22,830 harmonic functions, so we could write that our function is given by a sound function with the first 142 00:13:22,830 --> 00:13:28,410 ion frequency, plus a sound function with the second ion frequency plus assigned function with the 143 00:13:28,410 --> 00:13:29,760 third ion frequency. 144 00:13:30,600 --> 00:13:36,180 But of course, it's not so easy because all of these sine functions have different amplitudes and different 145 00:13:36,180 --> 00:13:37,080 face factors. 146 00:13:37,530 --> 00:13:43,530 So this means we have six different parameters that we first had to figure out if we want to fit this 147 00:13:43,530 --> 00:13:44,280 to the data. 148 00:13:45,090 --> 00:13:48,390 And actually, this is what we want to do in the next lecture. 149 00:13:49,560 --> 00:13:55,200 Just one more thing that I want to show you is that if I just rerun the whole notebook, we start with 150 00:13:55,200 --> 00:13:56,800 different starting conditions. 151 00:13:57,420 --> 00:14:02,040 And so this will give rise to different intensity spectra each of the time. 152 00:14:02,670 --> 00:14:09,750 For example, now you see that in all three cases, the one point four case does not really have a peak, 153 00:14:09,750 --> 00:14:12,030 except here, maybe a very, very small one. 154 00:14:12,870 --> 00:14:18,790 So let's do it once again, we have once again different starting conditions, and you see the amplitudes 155 00:14:18,810 --> 00:14:22,030 of these individual frequencies can be very, very different. 156 00:14:22,050 --> 00:14:27,390 For example, now it is even the largest peak, even though previously it wasn't there at all. 157 00:14:28,350 --> 00:14:35,220 But you always see the same thing that you do not see this peak for the second oscillator and that the 158 00:14:35,220 --> 00:14:37,890 first and the third one always looked the same. 159 00:14:38,700 --> 00:14:45,330 But the intensity of the peaks will always be different, and it depends strongly on the starting condition.