1 00:00:03,640 --> 00:00:08,020 In this video, we're going to have a look at different ways of how we can go about calculating the 2 00:00:08,020 --> 00:00:10,680 Giacobbe and matrixes that are required for the filter. 3 00:00:11,500 --> 00:00:18,340 So previously we have calculated the Jacoby Matrixes analytically from equations using differentiation, 4 00:00:18,520 --> 00:00:20,720 and we can see that this can be tedious. 5 00:00:21,040 --> 00:00:26,650 We take the process model, measurement model, and then we have to differentiate the model with respect 6 00:00:26,650 --> 00:00:30,210 to all the different inputs for the state, for the noise, for any inputs. 7 00:00:30,850 --> 00:00:34,760 And this can be quite complex for very nonlinear systems. 8 00:00:35,920 --> 00:00:39,190 There's also another way we can calculate the Jacobean matrix. 9 00:00:39,670 --> 00:00:44,860 It is possible to calculate this Jacobean matrix numerically straight from the model itself. 10 00:00:45,880 --> 00:00:52,630 And we can do this using numerical differentiation to get an idea of what numerical differentiation 11 00:00:52,630 --> 00:00:53,000 is. 12 00:00:53,290 --> 00:00:58,750 Imagine that we have a function F X and we want to differentiate the function with respect to X.. 13 00:00:59,470 --> 00:01:06,160 So this is basically saying we want to look at how the function output f changes with changes in X and 14 00:01:06,160 --> 00:01:09,030 we can actually evaluate this using first principles. 15 00:01:09,430 --> 00:01:15,640 So using first principles, if we have the function F evaluated at our initial condition or where we 16 00:01:15,640 --> 00:01:22,060 want to, linear is a system about plus a little perturbed amount subtracted away from the function 17 00:01:22,060 --> 00:01:26,920 evaluated at the initial condition, divided up by the D x amount. 18 00:01:27,190 --> 00:01:29,000 We basically end up with this function here. 19 00:01:29,410 --> 00:01:37,120 So this function looks at how F changes with changes in X, so basically we perturb X by a little amount 20 00:01:37,120 --> 00:01:44,050 by the X and then we subtract off the initial condition so we get a change in our F and then we divide 21 00:01:44,050 --> 00:01:46,450 it by the amount that we changed X. 22 00:01:47,050 --> 00:01:52,020 And this is basically differentiation as the Delta X goes towards zero. 23 00:01:52,630 --> 00:01:59,020 But when we do this numerically, we apply a non a small non-zero value and then calculate these functions 24 00:01:59,020 --> 00:02:00,220 here and do the division. 25 00:02:00,700 --> 00:02:05,590 And this will give us a numerical approximation of the derivative of the function F. 26 00:02:07,880 --> 00:02:13,520 So we can actually have a look at this using some numbers, so let's say we have a function F X is equal 27 00:02:13,520 --> 00:02:15,600 to two X, so we put a one in here. 28 00:02:15,650 --> 00:02:17,440 It's going to give us to put a two. 29 00:02:17,450 --> 00:02:19,540 It's going to give us four and so forth. 30 00:02:19,550 --> 00:02:23,090 And we want to differentiate this function with respect to X.. 31 00:02:24,320 --> 00:02:29,920 So what we want to do is that we can define Adella X, so let's define a small perturbed amount of point 32 00:02:29,990 --> 00:02:30,350 one. 33 00:02:30,800 --> 00:02:37,630 And we want to carry out this whole process around our initial condition of, say, X equals one. 34 00:02:38,000 --> 00:02:43,820 So we want to work out the derivative of this function around the value one. 35 00:02:44,540 --> 00:02:49,070 So what we can do is we can basically take these equations and these numbers, put them into this equation 36 00:02:49,070 --> 00:02:58,040 up here and evaluate it so we can substitute in if it's not plus Delta X, our function F X is just 37 00:02:58,040 --> 00:03:04,310 two X, so we get two times the X, which is one because we're evaluating it around one plus our total 38 00:03:04,310 --> 00:03:04,580 amount. 39 00:03:04,580 --> 00:03:07,900 D.T. sorry, Dex, giving us point one. 40 00:03:08,330 --> 00:03:16,120 So we get two times one plus point one and then we evaluate the function F at our initial condition. 41 00:03:16,130 --> 00:03:16,710 Excellent. 42 00:03:16,730 --> 00:03:24,650 So we put in our one into our two X, so this is our two XY and then we divide it by the perturbed amount 43 00:03:24,680 --> 00:03:25,850 which is point one. 44 00:03:26,880 --> 00:03:33,750 So they go through and do the calculation, we end up with their derivative equal to and if we know 45 00:03:33,900 --> 00:03:38,310 from numerical differentiation, if we differentiate this function here, the two X with respect to 46 00:03:38,310 --> 00:03:41,280 X, the X drops away, I mean, just off of two. 47 00:03:41,280 --> 00:03:45,990 So I can see that in this case, this is the numerical approximation of the derivative and it's going 48 00:03:45,990 --> 00:03:48,280 to be exactly equal to the real derivative. 49 00:03:49,350 --> 00:03:55,020 Now, this is a very benign example because the function to X is a linear system and finding the derivative 50 00:03:55,020 --> 00:03:55,730 is very simple. 51 00:03:56,190 --> 00:03:59,370 But this approximation holds for nonlinear systems as well. 52 00:04:03,040 --> 00:04:09,160 So then how can this process of numerical differentiation help us to numerically calculate the Jacobean? 53 00:04:10,450 --> 00:04:15,230 Well, let's say we give them a function Y is equal to F X you. 54 00:04:15,250 --> 00:04:17,530 So this is a function F and we have two inputs. 55 00:04:17,540 --> 00:04:22,660 We have an input X and an input you and we get an output of this whole function Y. 56 00:04:23,830 --> 00:04:31,630 As also say that the inputs X you and the output Y are also vector functions, so our X is an X vector. 57 00:04:31,660 --> 00:04:34,870 We have X states from one or way up to end. 58 00:04:35,410 --> 00:04:41,980 Our U vector is going to be a vector of all of you components from one to L, and the output is going 59 00:04:41,980 --> 00:04:43,840 to be output from one to M. 60 00:04:46,360 --> 00:04:48,190 So let's define some input's. 61 00:04:49,210 --> 00:04:55,660 So let's define X not being the initial condition that we want to calculate the UK about, so we store 62 00:04:55,690 --> 00:04:58,120 the initial value of the X Factor. 63 00:04:58,900 --> 00:05:05,760 Now we're going to define the X iveta, which is just the sum of the initial X Factor, plus this POTO 64 00:05:05,770 --> 00:05:13,060 factor here and this term vector is just going to have the element in this vector, the Delta X and 65 00:05:13,060 --> 00:05:14,970 all the other elements equal to zero. 66 00:05:15,340 --> 00:05:23,150 So it just means we have the vector which has the element of X not perturbed by an amount Delta X. 67 00:05:23,620 --> 00:05:25,390 So all this equation is showing here. 68 00:05:25,390 --> 00:05:33,370 If we evaluate our Delta X, J and J and I are equal to each other, we set the value to be Delta X, 69 00:05:33,370 --> 00:05:34,780 otherwise we will be zero. 70 00:05:35,230 --> 00:05:41,830 So if this was X1, then this vector over here is just going to be our Delta X because X1 is going to 71 00:05:41,830 --> 00:05:47,800 be equal to Delta X when I enjoy equal one and all the other times here going to be zero zero zero zero. 72 00:05:48,400 --> 00:05:54,550 Likewise, if this was X three, this vector over here is going to be zero zero dollars zero zero zero 73 00:05:54,550 --> 00:05:55,720 zero all the way up to X. 74 00:05:55,720 --> 00:06:04,720 And so this is just the probation of this vector X with the element perturbed by the value Delta X. 75 00:06:05,940 --> 00:06:09,280 We can go about and define the same thing for the year. 76 00:06:09,810 --> 00:06:12,350 So basically, since we have a function here, it takes two inputs. 77 00:06:12,360 --> 00:06:17,540 We have to define the initial condition or the linear ization point for both these inputs. 78 00:06:17,880 --> 00:06:20,160 And in this case is going to be our, you know. 79 00:06:21,320 --> 00:06:27,980 Now, we can also perturb the Uvalda in the same way, so basically we define our UI is just going to 80 00:06:27,980 --> 00:06:34,970 be our, you know, vector and then a perturbed vector, which has the element inside this vector set 81 00:06:34,970 --> 00:06:38,900 to Delta, you and all the other elements set to zero. 82 00:06:42,600 --> 00:06:45,450 Now, using these inputs, we can define some outputs. 83 00:06:45,940 --> 00:06:51,990 We're going to define why not just going to be the value of this function he evaluated at X not and 84 00:06:51,990 --> 00:06:52,470 you know. 85 00:06:53,460 --> 00:07:02,190 We're also we're also going to define our why EXI is going to be the function evaluated at Aleksi and 86 00:07:02,190 --> 00:07:07,720 you know, so this is evaluating the function with the element of expert. 87 00:07:08,100 --> 00:07:13,890 And we can show you the same thing for you so we can have a wide UI, which is evaluating the function 88 00:07:13,890 --> 00:07:17,430 with the element of the you input vector perturbed. 89 00:07:18,120 --> 00:07:22,830 So now that we define these different parameters, we can look at doing the actual Jacobean calculation 90 00:07:22,830 --> 00:07:23,880 and forming the matrix. 91 00:07:25,630 --> 00:07:31,450 So we can approximate the Jakobsson of the system evaluated at these liberalization point as being one 92 00:07:31,450 --> 00:07:39,010 over Deltour X, and then the columns of the recovery matrix are just going to be Y, X, one minus 93 00:07:39,010 --> 00:07:39,550 Y not. 94 00:07:39,790 --> 00:07:46,090 And then it's going to be the Y x2 minus Y, not all the way up to Y and explain why not. 95 00:07:46,510 --> 00:07:52,390 So basically we're going to form the columns of the covid matrixes just from differences of these vectors 96 00:07:52,390 --> 00:07:56,490 up here, and then we're going to scale it all by Alberto to mount the X. 97 00:07:57,520 --> 00:08:02,040 So this is just doing what we did on the first slide with the scalar example. 98 00:08:02,230 --> 00:08:04,440 But we're going to do it now in vector form. 99 00:08:05,710 --> 00:08:08,910 And when we do, this will end up with this describing matrix here. 100 00:08:08,920 --> 00:08:15,910 So we're going to have the Jacobean and outputs because our Y matrix is going to be of size M and it's 101 00:08:15,910 --> 00:08:22,060 going to have an input because our input function, Àlex Vector is going to be of size N. 102 00:08:23,880 --> 00:08:26,190 So this is going to be I am by N Matrix. 103 00:08:27,480 --> 00:08:31,240 We can do the same thing for the Caribbean with respect to the input. 104 00:08:31,680 --> 00:08:35,130 So now we're going to divide it by one over our daughter, you. 105 00:08:35,520 --> 00:08:41,630 And instead of using our X1 here, we're going to be using our you want you to all the way up to a year 106 00:08:41,640 --> 00:08:41,890 end. 107 00:08:42,270 --> 00:08:48,960 So basically, we're looking at how the output of the function changes with so probations in the U vector 108 00:08:48,960 --> 00:08:51,760 instead of small probation's in the X Factor. 109 00:08:52,770 --> 00:08:54,420 And when we do this, we're going to get out. 110 00:08:54,420 --> 00:08:56,110 You carry a matrix with respect to you. 111 00:08:56,130 --> 00:09:04,260 And again, it's going to have a output size of N by El where M is the number of outputs we have. 112 00:09:04,500 --> 00:09:13,470 So that's the size of the Y vector and is going to be of number of columns l where L is the size of 113 00:09:13,470 --> 00:09:15,150 our input vector U. 114 00:09:19,430 --> 00:09:25,010 So more concretely, let's actually have a look at how this algorithm works in general and how we can 115 00:09:25,010 --> 00:09:25,770 actually define it. 116 00:09:26,390 --> 00:09:29,240 So let's have a look at the state education algorithm. 117 00:09:29,270 --> 00:09:35,660 Now, this is the calculation of the Giacobbe matrix or the function F respect to X for the function 118 00:09:35,660 --> 00:09:37,400 given by Y is a Yoda. 119 00:09:37,400 --> 00:09:38,570 If you. 120 00:09:39,850 --> 00:09:44,470 Given the initial conditions, it's not and, you know, so these are the conditions that we want to 121 00:09:44,470 --> 00:09:49,330 linearity the system about or calculate the Jacobin of the nominee system about. 122 00:09:51,210 --> 00:09:56,930 The first step in the process is to calculate the appropriate amount X, so we're going to set Delta 123 00:09:56,940 --> 00:09:59,090 X to be a small numerical value. 124 00:09:59,880 --> 00:10:02,510 The next step is that we want to calculate the initial state. 125 00:10:02,880 --> 00:10:11,460 So we take the linear Azanian position, we put it into a function F and we calculate why not now for 126 00:10:11,460 --> 00:10:17,700 each element in the statement X, so we know we have x1 x2 all the way up to X and we want to perform 127 00:10:17,700 --> 00:10:19,100 the following operations. 128 00:10:19,530 --> 00:10:26,700 We first want to calculate the vector, so we want to calculate x y, which is just going to be Alisher, 129 00:10:27,120 --> 00:10:32,400 which is just going to be our initial condition, plus the better vector which has the element in this 130 00:10:32,400 --> 00:10:35,450 vector perturbed and all the other elements equal to zero. 131 00:10:35,490 --> 00:10:37,590 That's what this equation here is explaining. 132 00:10:38,220 --> 00:10:42,640 Once we have the perturbed input vector, then we want to calculate the perturbed output. 133 00:10:42,660 --> 00:10:49,260 So that's basically substituting in our perturbed X Factor into the process model or into the function 134 00:10:49,260 --> 00:10:52,140 and calculating output of the perturbed input. 135 00:10:53,490 --> 00:10:58,260 Then we can calculate the difference between the vectors, so the perturbed output vector minus the 136 00:10:58,260 --> 00:11:04,320 initial output vector divided by the total amount is going to give us the eighth column in the Giacobbe 137 00:11:04,320 --> 00:11:10,410 matrix so we can just go through and do this multiple times, building up the metrics and we're going 138 00:11:10,410 --> 00:11:16,040 to loop through end times until we get all the end input states for X. 139 00:11:17,930 --> 00:11:20,570 So there's nothing too complicated going on with this algorithm here. 140 00:11:20,590 --> 00:11:25,460 We just want to go through, build up the Jacoby and Matrix column by column, and each time we do that, 141 00:11:25,460 --> 00:11:30,620 we just perturb X by a tiny bit and see how that affects the output and then calculate the difference 142 00:11:30,620 --> 00:11:33,860 of this output and scale it by a bit perturbed amount. 143 00:11:34,900 --> 00:11:42,010 But we can also do this for any of the other inputs inside the function, so this function F X has actually 144 00:11:42,010 --> 00:11:44,550 two inputs, has an X factor and a new vector. 145 00:11:45,250 --> 00:11:48,610 So we perturb the X Factor to get the extra came in. 146 00:11:49,360 --> 00:11:51,870 We can do the same thing for the input you. 147 00:11:51,880 --> 00:11:56,980 So if you want to work out the Giacobbe in respect to the other input, you we do basically the same 148 00:11:56,980 --> 00:12:00,610 thing, except instead of perturbing X, we're now perturbing you. 149 00:12:00,620 --> 00:12:06,820 So we have a defined observed amount you and then we want to go through for every element inside the 150 00:12:06,820 --> 00:12:08,010 input vector you. 151 00:12:08,290 --> 00:12:14,590 So for one all the way up to L we want to calculate the perturbed you vector substitutability into the 152 00:12:14,590 --> 00:12:17,260 function and hold X constant this time. 153 00:12:17,260 --> 00:12:24,280 But now we're perturbing you work out the output and then calculate the Caribbean for and calculate 154 00:12:24,280 --> 00:12:31,180 the Jacobin by building up column by column with respect to the differences in the output for you. 155 00:12:31,180 --> 00:12:32,260 Perturbations. 156 00:12:35,060 --> 00:12:40,520 The calculating the GDP matrix numerically can be very useful, instead of sitting down and having to 157 00:12:40,520 --> 00:12:46,760 go through all the mathematical operations to differentiate your process model, you can do it numerically. 158 00:12:46,760 --> 00:12:52,760 And this becomes really handy, especially if you have a large system of like 50 states or even 12 states 159 00:12:53,090 --> 00:13:00,170 trying to work out the analytical Jacquemin metrics can be very error prone and take a lot of time while 160 00:13:00,290 --> 00:13:03,320 just using a numerical algorithm is perfectly acceptable. 161 00:13:03,500 --> 00:13:09,560 And it gives you results that are close enough to the true analytical solution that the output of the 162 00:13:09,560 --> 00:13:13,970 filter would be indifferent to whether you use the approximation or the true Jacobean. 163 00:13:14,600 --> 00:13:19,520 So hopefully you have an idea of how we can go about calculating the Giacobbe metrics numerically and 164 00:13:19,520 --> 00:13:22,300 hopefully you can see how it can have some advantages. 165 00:13:22,550 --> 00:13:27,470 It stops you from doing having to do all the calculations by hand and it can save you from making small 166 00:13:27,470 --> 00:13:29,330 errors which are very hard to track down.