1 00:00:03,770 --> 00:00:09,140 So now let's actually put some numbers behind it and look at a real world example so we can understand 2 00:00:09,140 --> 00:00:09,950 what's going on. 3 00:00:11,060 --> 00:00:14,090 So let's consider the vehicle process model that we've been using. 4 00:00:14,090 --> 00:00:15,940 And that's basically our process model. 5 00:00:15,950 --> 00:00:22,370 If we take in the previous state and current controls and calculate the new state for the current time 6 00:00:22,370 --> 00:00:22,710 step. 7 00:00:23,270 --> 00:00:30,140 So in the two day vehicle process model, we have our state vector being our velocity heading position 8 00:00:30,140 --> 00:00:36,080 in the X and position and Y, our inputs to the same process model are going to be our accelerations. 9 00:00:36,080 --> 00:00:40,580 And you're right, and these are the equations that we have been using. 10 00:00:41,360 --> 00:00:46,910 So we're going to go through an example where we use this to calculate the Jacobean of this process 11 00:00:46,910 --> 00:00:47,510 model here. 12 00:00:48,680 --> 00:00:54,410 So first off is that we want to define the previous state that we want to linear the system about. 13 00:00:54,410 --> 00:01:00,680 And we're going to use this state vector here so we can have a velocity equals 2.5 hour heading is going 14 00:01:00,680 --> 00:01:03,530 to be our point seven or seven radians. 15 00:01:03,710 --> 00:01:05,470 Our exposition is going to be 10. 16 00:01:05,720 --> 00:01:07,970 A Y position is going to be negative five. 17 00:01:09,240 --> 00:01:15,030 Now, for our current input, we're going to assume that we're linear, rising at around a acceleration 18 00:01:15,030 --> 00:01:20,490 value of one point five and your rate of negative point to five. 19 00:01:23,040 --> 00:01:29,070 We're also going to need to define a few parameters, we're going to define a date value of point one, 20 00:01:29,520 --> 00:01:34,370 and when we're calculating the Jacoby numerically, we're going to be using a Delta X. 21 00:01:34,380 --> 00:01:37,650 So this is the perturbed amount that we want to be to the X states. 22 00:01:37,650 --> 00:01:43,680 It's going to be our point zero one and we're going to do the same value for our total amount for the 23 00:01:43,680 --> 00:01:45,690 input 0.01. 24 00:01:47,610 --> 00:01:54,030 So this state here becomes the initial condition that we want to lean your eyes for The X Factor, and 25 00:01:54,030 --> 00:01:59,270 this input here is going to become our initial condition for the input that we want to see in your eyes 26 00:01:59,280 --> 00:02:00,380 and system about as well. 27 00:02:01,630 --> 00:02:07,270 So the first step in the process is to calculate the output for the militarization point, so that's 28 00:02:07,270 --> 00:02:13,330 basically what we put in our external and, you know, values into the function to evaluate what it 29 00:02:13,330 --> 00:02:15,200 would be in this condition here. 30 00:02:15,670 --> 00:02:20,890 So when we do that numerically, we take these equations here, substitute in these values here, and 31 00:02:20,890 --> 00:02:23,510 we end up with this series of equations here. 32 00:02:24,190 --> 00:02:28,620 So basically, we're taking out like velocity two point five here. 33 00:02:28,630 --> 00:02:34,290 Our data point one times our acceleration amount one point five. 34 00:02:34,300 --> 00:02:37,490 And when we do the calculation, we're going to get our two point six five. 35 00:02:37,720 --> 00:02:43,300 And when you do the same thing for a heading and the position and the X and position and the Y and we 36 00:02:43,300 --> 00:02:45,310 end up with this Y, not vector. 37 00:02:46,660 --> 00:02:49,980 Now we also want to do it for our X1. 38 00:02:50,020 --> 00:02:56,530 So this is when we perturb the first element inside our vector and then we do the same process. 39 00:02:56,530 --> 00:03:02,270 We put the we put the vector into our process model and evaluate the output. 40 00:03:03,010 --> 00:03:09,400 So when we perturb the first state of velocity, we're going to be tabatabai out the X amount, our 41 00:03:09,400 --> 00:03:11,210 point zero one. 42 00:03:11,230 --> 00:03:18,150 So we now wherever we have velocity of 2.5, we're going to have 2.5 plus zero point zero one. 43 00:03:18,700 --> 00:03:24,550 So you can see here we had the velocity and we also perturb it down here because we have these velocity 44 00:03:24,550 --> 00:03:25,440 terms down here. 45 00:03:25,450 --> 00:03:27,500 This is we're also using the velocity here. 46 00:03:28,210 --> 00:03:33,190 So this time when we go through and do all these equations, we're going to get slightly different numbers 47 00:03:33,400 --> 00:03:35,700 for anything that's related to velocity. 48 00:03:35,740 --> 00:03:37,350 So I can see the top 10 here. 49 00:03:37,780 --> 00:03:43,090 We're going to get a slight change in velocity because now we return the WASSE and redone the calculations. 50 00:03:43,630 --> 00:03:46,360 The heading equation does not rely on velocity. 51 00:03:46,390 --> 00:03:47,850 So we're going to get the same value here. 52 00:03:48,550 --> 00:03:52,240 But Alex and Y positions do depend on velocity. 53 00:03:52,270 --> 00:03:55,560 So we're going to slightly different answers than we would have up here. 54 00:03:57,350 --> 00:04:03,110 So taking the values that we've just calculated, we can evaluate this term or the Jacobean, so we've 55 00:04:03,110 --> 00:04:11,900 calculated our Y at X one minus Y not so we can do this calculation so we can divide by Abbotabad amount 56 00:04:11,900 --> 00:04:15,980 outpoint zero one three one divided by outpoint zero one. 57 00:04:16,430 --> 00:04:22,160 We look at the differences between our Petare vector output and our non perturbed output. 58 00:04:22,380 --> 00:04:26,870 So the difference between the two states here and we end up with this difference vector. 59 00:04:30,320 --> 00:04:35,630 Now, this difference, Victor, actually makes up the first column of our Jacobean so we can take this 60 00:04:35,990 --> 00:04:38,360 factor here and put it into the first column. 61 00:04:38,360 --> 00:04:40,190 So that's this term here. 62 00:04:40,610 --> 00:04:44,210 So this is how we get the first column of the Giacobbe matrix. 63 00:04:46,190 --> 00:04:53,960 The next step, we can do the same thing now, but we can evaluate a why X two, so why X is just going 64 00:04:53,960 --> 00:04:54,800 to be the same thing. 65 00:04:54,800 --> 00:05:00,920 But this time the input for The X Factor is going to be X two, which means we're going to perturb the 66 00:05:00,920 --> 00:05:02,420 second element of X. 67 00:05:02,810 --> 00:05:08,630 So here where we have the heading way, the second element, the point seven zero seven, we're going 68 00:05:08,630 --> 00:05:13,910 to add on our Delta X of point zero one and redo the calculations. 69 00:05:14,330 --> 00:05:20,840 So wherever we have heading inside the equation, so wherever we have any value of heading, we want 70 00:05:20,840 --> 00:05:26,870 to take the original heading peterschmidt by a tiny bit same thing down here in here, and then calculate 71 00:05:26,870 --> 00:05:27,440 the result. 72 00:05:27,440 --> 00:05:31,650 And we're going to end up with these elements here inside the final vector for the upper. 73 00:05:31,910 --> 00:05:35,810 And you can see that any element that depends on heading is slightly different. 74 00:05:35,810 --> 00:05:37,370 So the numbers are going to be slightly different. 75 00:05:38,840 --> 00:05:43,790 Once we've calculated this, Victor, we can put it into this equation off the top here, and so we're 76 00:05:43,790 --> 00:05:45,770 calculating now the second column. 77 00:05:45,800 --> 00:05:50,270 So if we calculate the difference between the two vectors, we end up with this vector. 78 00:05:51,260 --> 00:05:58,130 And this factor is going to be the second column of the Caribbean so we can do the same thing, perturbing 79 00:05:58,130 --> 00:05:59,220 the third component. 80 00:05:59,240 --> 00:06:04,760 So now we're perturbing the exposition, putting it into our series of equations, recalculating it. 81 00:06:04,760 --> 00:06:07,790 And anything that depends on X is now going to be slightly different. 82 00:06:08,630 --> 00:06:14,480 We follow the same process of getting the difference of the vector, dividing it by our perturbed amount, 83 00:06:14,840 --> 00:06:20,970 and then that output becomes the wall, the third column inside the Matrix. 84 00:06:21,740 --> 00:06:25,190 And then lastly, we do the last step for our X4. 85 00:06:25,190 --> 00:06:26,710 So we perturb a lost state. 86 00:06:27,080 --> 00:06:31,810 So we perturbing the Y position by a small amount and redoing the calculations. 87 00:06:31,820 --> 00:06:34,280 And again, we get this set of equations here. 88 00:06:34,700 --> 00:06:38,030 Wherever we have y position, we want to perturb it. 89 00:06:38,210 --> 00:06:44,480 So down here we have our negative five plus a little perturbed amount and nothing else in this equation 90 00:06:44,480 --> 00:06:47,720 depends on Y, so that's the only element that's going to be perturbed. 91 00:06:48,080 --> 00:06:55,370 Redo the calculations, see what's changed, and then subtract it from the initial output to find out 92 00:06:55,640 --> 00:06:59,630 the final column in the Jacobean matrix as shown here. 93 00:07:00,530 --> 00:07:04,310 So this is how we go about building up the Giacobbe matrix step by step. 94 00:07:06,900 --> 00:07:11,620 So this is a numerical approximation of the analytical code, you cave in. 95 00:07:11,760 --> 00:07:17,660 So we've come up with this numerical approximation without Delta X to be point zero one. 96 00:07:17,940 --> 00:07:20,660 And these are the numerical values that we've just calculated. 97 00:07:20,970 --> 00:07:25,620 So we can say that this is a numerical approximation of the Jacobean matrix. 98 00:07:26,490 --> 00:07:32,100 We have also calculated the analytical solution in the past using differentiation, and we came up with 99 00:07:32,100 --> 00:07:33,880 this equation set here. 100 00:07:34,440 --> 00:07:40,620 So if we actually calculate the analytical solution by substituting in our parameters and states into 101 00:07:40,620 --> 00:07:46,710 this matrix here, our real solution for the analytical solution of the Jacobin is this matrix here. 102 00:07:47,280 --> 00:07:51,100 So if we compare the values here, we can see that they're very similar. 103 00:07:51,550 --> 00:07:55,170 They're only slightly difference in a few different decimal places. 104 00:07:56,860 --> 00:08:03,850 So how accurately we calculate the Jacoby Imageworks using the numerical approximation is going to depend 105 00:08:03,850 --> 00:08:06,940 on the size of our perturbed amount, the X. 106 00:08:08,650 --> 00:08:15,010 We want to ideally use a very small value of Deltour X, so in this case, if we redo the calculations 107 00:08:15,010 --> 00:08:20,980 but use a even smaller value of Delta X, you can find that the analytical solution is even closer to 108 00:08:20,980 --> 00:08:27,010 the numerical solution and is going to be pretty much exactly the same all the way down to the numerical 109 00:08:27,340 --> 00:08:29,170 precision of the computer. 110 00:08:31,810 --> 00:08:37,570 Now, unfortunately, we can't just put in the smallest number here that the computer can support because 111 00:08:37,750 --> 00:08:43,700 because of rounding errors inside the system, so we want to make sure we use a small enough donor X 112 00:08:43,720 --> 00:08:45,070 for the system we're using. 113 00:08:45,490 --> 00:08:49,870 If we use a large donor X, we're going to get a bad approximation of the Ukrainian. 114 00:08:50,230 --> 00:08:56,830 But anything around this order or smaller is usually a fairly good approximation of the Ukrainian matrix.