1 00:00:00,090 --> 00:00:04,770 OK, let's start with one of the simplest centralized control methods. 2 00:00:05,340 --> 00:00:12,450 This is simple, but yet are private yet better than previous decentralized control schemes where we 3 00:00:12,450 --> 00:00:16,380 didn't consider dynamic model of the robot manipulator. 4 00:00:17,100 --> 00:00:21,240 You can ask here why we don't utilize simple PD controller. 5 00:00:21,570 --> 00:00:28,500 As we have said before, the centralized control method doesn't take into account tools that are generated 6 00:00:28,500 --> 00:00:34,920 due to different effects in the robot dynamics so that it doesn't guarantee the achievement of the position, 7 00:00:34,920 --> 00:00:38,870 control, objective and gravity talks on coincided. 8 00:00:39,300 --> 00:00:45,810 So we cannot assure that trajectory will be tracked and checked, and desired position will be reached 9 00:00:46,050 --> 00:00:53,640 by just applying simply PD or PD controller because it doesn't take into account gravity time. 10 00:00:54,000 --> 00:01:01,680 Indeed, with simple PD controller, we assume that the final configuration gravity toric will be zero, 11 00:01:01,680 --> 00:01:02,880 which is not correct. 12 00:01:03,180 --> 00:01:09,510 However, with PD and gravity compensation, we will remove this constraint and we'll try to achieve 13 00:01:09,510 --> 00:01:14,250 asymptotic of stability even when gravity vector is not zero. 14 00:01:14,640 --> 00:01:21,150 So as we have said before, our goal is to track the given trajectory with asymptotic, asymptotically 15 00:01:21,150 --> 00:01:22,800 stable aerodynamics. 16 00:01:23,100 --> 00:01:30,370 So we want to tell, though, which is position error and QTL, the DOT, which is velocity error, 17 00:01:30,660 --> 00:01:37,530 the conversion to zero and becomes zero at the end before jumping into the developing of control algorithm. 18 00:01:37,530 --> 00:01:39,960 Let's analyze one important motion. 19 00:01:40,170 --> 00:01:43,650 Let's analyze the steady state of the dynamic system. 20 00:01:43,650 --> 00:01:48,780 As we have said, we want to till the and to tell the DOT to be zero. 21 00:01:49,050 --> 00:01:53,790 So we can achieve desired velocity and configuration we want. 22 00:01:54,240 --> 00:01:57,120 Here is a dynamic model of the robot manipulator. 23 00:01:57,390 --> 00:02:06,240 We can write this equation in steady state as in this form as you can see to become desired and could 24 00:02:06,380 --> 00:02:08,310 becomes kudo desired. 25 00:02:08,550 --> 00:02:12,060 And also gravity term has been compensated fully. 26 00:02:12,390 --> 00:02:14,280 This is our purpose, isn't it? 27 00:02:14,460 --> 00:02:21,540 Because the name of the controller is PD and gravity compensation name and gravity is compensated for. 28 00:02:22,350 --> 00:02:29,030 So in order to achieve this kind of steady state C on, some conditions have to be met from above equation. 29 00:02:29,050 --> 00:02:35,580 You can see that in order the sum to be zero, all of the components have to be zero separately. 30 00:02:35,640 --> 00:02:42,150 Y as m metrics or inertia matrix is always positive, doesn't it surely kill? 31 00:02:42,150 --> 00:02:44,610 The double thought has to be zero. 32 00:02:44,910 --> 00:02:47,100 Otherwise, this term will not be zero. 33 00:02:47,340 --> 00:02:54,570 Moreover, in order other terms to be zero Q, the Dot has to be zero because when it becomes zero, 34 00:02:54,840 --> 00:03:02,490 C Matrix becomes zero automatically and the sum and the search term also becomes zero automatically 35 00:03:02,940 --> 00:03:04,890 from these two constraints. 36 00:03:05,130 --> 00:03:15,120 We did use that in order to achieve both steady state condition Q de la Q, the name the desired configuration 37 00:03:15,360 --> 00:03:17,040 has to be constant. 38 00:03:17,430 --> 00:03:25,020 Don't forget that if the is constant, then Q Tilde Dot will become just minus Q Dot because cutely 39 00:03:25,020 --> 00:03:27,180 dot will become zero. 40 00:03:27,750 --> 00:03:33,960 OK, now we can deep dive or become active or dive deep. 41 00:03:34,130 --> 00:03:40,740 Yeah, let's first determine state vector for aerodynamics. 42 00:03:41,490 --> 00:03:49,440 We want aerodynamics to BS Asymptotically stable, then the second worst to zero because if C converts 43 00:03:49,440 --> 00:03:52,740 to zero, then Q tilde and the Q becomes zero. 44 00:03:53,100 --> 00:03:56,340 So if we achieve desired steady state condition. 45 00:03:59,540 --> 00:04:05,720 In order to develop a control algorithm that will make ever dynamics asymptotically stable, we will 46 00:04:05,720 --> 00:04:07,970 utilize in the optimal analysis method. 47 00:04:08,600 --> 00:04:15,500 So we will first choose short the validly optimal function and it's given in this way, Cape Matrix 48 00:04:15,500 --> 00:04:18,140 is positive, definite and symmetric metrics. 49 00:04:18,620 --> 00:04:24,380 If you look carefully, you will see that the first term of the neptunium function is in the form of 50 00:04:24,420 --> 00:04:28,010 dynamic energy, while the second is potential energy. 51 00:04:28,460 --> 00:04:41,210 As with potential energy as kinetic and potential energies are always positive definite, then the optimal 52 00:04:41,210 --> 00:04:47,270 function will also be positive, definite except in the origin, namely when quiet and security that 53 00:04:47,270 --> 00:04:48,140 becomes zero. 54 00:04:48,590 --> 00:04:51,830 This proves that we have little the Apollo function. 55 00:04:52,250 --> 00:04:59,960 Also note that we have chosen the optimal function such that it includes Q and Q tilde variables. 56 00:05:00,290 --> 00:05:07,340 So when the steady state has been reached, namely when vetoed becomes zero, they will become zero. 57 00:05:07,340 --> 00:05:08,810 Also perfect. 58 00:05:09,110 --> 00:05:14,690 Now let's analyze the derivative of the optimal function in order to see whether it is negative. 59 00:05:14,690 --> 00:05:18,710 Definite so error dynamics is asymptotically stable or not. 60 00:05:19,430 --> 00:05:26,360 I will start with the derivative of the kinetic energy because I want to show the calculation at this 61 00:05:26,360 --> 00:05:26,660 point. 62 00:05:27,050 --> 00:05:33,950 As you can see first, we have applied simple derivative of summation role as an is symmetric and positive. 63 00:05:33,950 --> 00:05:41,150 Definite it's transpose will be equal to itself, so we can write equation in this way also without 64 00:05:41,150 --> 00:05:42,170 changing the result. 65 00:05:42,500 --> 00:05:50,540 Now let's analyze the dimension of green products as you can see the product of Q, the transpose inertia 66 00:05:50,540 --> 00:05:56,300 matrix and Q table both will be grown in the Q. 67 00:05:56,990 --> 00:06:03,820 Excuse me, as you can see the product of Q transpose Enoshima please and could double. 68 00:06:03,830 --> 00:06:06,140 Both will be one by one named Scarlet. 69 00:06:06,140 --> 00:06:12,610 While you saw two green terms are equal to each other and we can write them in this way. 70 00:06:12,620 --> 00:06:20,240 Also so derivative only a form of function can be written like that easily. 71 00:06:22,430 --> 00:06:29,210 If you find into to double talk from Standard Dynamics equation and plug in this equation, you will 72 00:06:29,210 --> 00:06:36,620 obtain that equation where entered minus two C will be equal to zero. 73 00:06:37,550 --> 00:06:40,820 C matrix is constructed by Christoffer Symbols. 74 00:06:41,390 --> 00:06:44,840 So finally, we can write V Dot in this form. 75 00:06:46,250 --> 00:06:49,070 As you can see, first is negative definite. 76 00:06:49,070 --> 00:06:50,720 But what about the second term? 77 00:06:51,860 --> 00:06:52,580 We don't know. 78 00:06:52,940 --> 00:06:54,610 We know that we want. 79 00:06:54,610 --> 00:06:56,990 We thought to be negative, definite. 80 00:06:57,290 --> 00:07:01,280 So what we can do is to eliminate the term with unknown. 81 00:07:01,280 --> 00:07:05,150 So let's see the Vidot again. 82 00:07:05,330 --> 00:07:11,030 If we choose to name the input control in this way, then vetoed will surely be negative. 83 00:07:11,030 --> 00:07:16,010 Definite except Kyoto becomes zero except when you don't become zero. 84 00:07:16,880 --> 00:07:23,840 Saw a birth control method is nothing but linear PD control with compensation of nonlinear gravity terms. 85 00:07:24,200 --> 00:07:28,610 So the name of the control is P D and the gravity compensation. 86 00:07:28,940 --> 00:07:36,080 Here, I want to note one thing if we would choose input control TAL as on the P control and the gravity 87 00:07:36,080 --> 00:07:39,770 compensation, it would also make the negative definite. 88 00:07:40,100 --> 00:07:43,310 But why we have added additional derivative term. 89 00:07:43,760 --> 00:07:48,380 This is because robots are made with as minimum friction as possible. 90 00:07:49,250 --> 00:07:54,800 Mechanical engineers tries to decrease friction in robot components as much as possible. 91 00:07:54,830 --> 00:08:01,370 However, on the other side, friction helps to improve systems time, response and convergence the 92 00:08:01,400 --> 00:08:02,180 equilibrium. 93 00:08:02,570 --> 00:08:09,290 So we add additional derivative term to make convergence to the equilibrium faster, namely improve 94 00:08:09,300 --> 00:08:10,190 time response. 95 00:08:10,640 --> 00:08:13,630 You can see this by comparison. 96 00:08:13,640 --> 00:08:22,700 We by comparing video terms when Paul is chosen as PD and gravity compensation and when chosen as P 97 00:08:22,700 --> 00:08:24,650 Control V the correct compensation. 98 00:08:25,100 --> 00:08:32,420 Another thing to note is, as we have said before, we were both Q2 and Q3 of the DOT to be zero when 99 00:08:32,420 --> 00:08:33,860 the Dot becomes zero. 100 00:08:34,400 --> 00:08:36,950 So to get the desired steady state at the end. 101 00:08:37,250 --> 00:08:44,480 But if you analyze vetoed in above equation, you will see that it contains only CU adult term, namely 102 00:08:44,480 --> 00:08:48,410 only cumulative becomes zero when we don't become zero. 103 00:08:48,830 --> 00:08:53,180 But what about the CU til the term then won't be reached. 104 00:08:53,180 --> 00:08:57,980 The desired configuration by PD and part of the compensation control lets on the. 105 00:08:58,230 --> 00:09:04,290 Still steady state of the system to learn that here I have written dynamics equation again with Tom 106 00:09:04,290 --> 00:09:11,370 replaced by P.D. and the gravity compensation control we have formed before, instead of state kudos 107 00:09:11,660 --> 00:09:19,530 becomes zero and also double dot because robot stops moving and gravity terms will take each other out 108 00:09:19,530 --> 00:09:22,290 and we will end up with that equality. 109 00:09:22,710 --> 00:09:30,880 You can see from here that Churchill will become zero at steady state, so we will not only obtain cute 110 00:09:30,900 --> 00:09:36,590 not equals to zero, but also be able reach desired configuration we want. 111 00:09:36,600 --> 00:09:37,470 That's perfect. 112 00:09:38,040 --> 00:09:45,750 However, keep in mind that this global asymptotic stability belong only or the ideal case where we 113 00:09:45,750 --> 00:09:49,410 can predict the value of gravity took victor exactly. 114 00:09:49,740 --> 00:09:52,530 And so compensation becomes exact. 115 00:09:52,950 --> 00:09:59,370 In practice, we cannot calculate good over the term exactly due to uncertainties in the dynamic parameters 116 00:09:59,370 --> 00:10:00,390 of robot model. 117 00:10:00,690 --> 00:10:07,800 So compensation will be partial, not complete, and we will not be able to get asymptotic stability 118 00:10:07,980 --> 00:10:09,750 but simple stability. 119 00:10:10,140 --> 00:10:17,780 As long as KP and KD mattresses are positive definite, we will see this concept more clearly during 120 00:10:17,790 --> 00:10:19,320 robust control techniques. 121 00:10:21,120 --> 00:10:26,490 And here is the clear diagram of the PD and gravity compensation control technique. 122 00:10:27,660 --> 00:10:35,730 As you can see, the red part is PD control to stabilize aerodynamics and gravity compensation with 123 00:10:35,730 --> 00:10:41,370 green power to provide, as we have said, gravity compensation in steady state.