<ul><li><p>IntroductionToRobotics-Lecture15 </p><p>Instructor (Oussama Khatib):Okay. Okay. Lets get started. So todays video segment is about tactile sensing. Now, I wonder what is difficult about building tactile sensors; anyone has an idea? So what is the problem with building a tactile sensor? Oh, you used to see the video first, okay. So, yeah. </p><p>Student:Do you need functions to be able to, I mean, do you need a perturbation to be able to see what youre touching sometimes? </p><p>Instructor (Oussama Khatib):Well, yeah, sometimes you, I mean, a human tactile sensing is amazing. So you have the static information, so if you grab something, now the whole surface is in contact, and you can determine the shape, right? So what does it mean in term of, like, designing a tactile sensor, just if you think about the static case? </p><p>Student:Its soft, malleable. </p><p>Instructor (Oussama Khatib):Well, you need some softness in the thing you are putting. Then you need to take this whole information, what kind of resolution do you need, if you are touching to feel the edge? You need a lot of pixels, right? So how can you take this information and first of all, how you determine that information; what kind of procedure do you yes? </p><p>Student:Well, theres an element of pressure, like, how hard youre the average how are you touching on all these different things. </p><p>Instructor (Oussama Khatib):Okay. So you can imagine, maybe, a sort of resistive or capacitive sensor that will deflect a little bit and give you that information. How many of those you would need? You need, sort of, an array, right? So how large, like, lets say this is the end of factor. Im trying to see if you did that problem youre going to have a lot of information here, and you need to take it back, and you have a lot of wires; you have a matrix, and youre going to have a lot of, basically, information to transmit. So, the design of tactile sensors being this problem of how we can put enough sensors, and how we can extract this information and take it back. So these guys came up with an interesting idea; here it is. The light, please. [Video]: </p><p>A novel tactile sensor using optical phenomenon was developed. In the tactile sensors shown here, light is injected at the edge of an optical wave guide made of transparent material and covered by an elastic rubber cover. There is clearance between the cover and the wave guide. The injected light maintains total internal reflection at the surface of the wave guide and is enclosed within it. When an object makes contact with it, the rubber cover depresses and touches the wave guide. Scattered light arises at the point of contact due to the change of the reflection condition. Such tactile information can be converted into a visual image. </p></li><li><p>Using this principle, a prototype finger-shaped tactile sensor with a hemispherical surface was developed. A CCD camera is installed inside the wave guide to detect scattered light arising at the contact location on the sensors surface. The image from the CCD camera is sent to the computer, and the location of the scattered light is determined by the image processing software. Using this information, the objects point of contact on the sensors surface can be calculated. </p><p>To improve the size and the operational speed of the sensor, a miniaturized version was developed. The hemispherical wave guide with cover, the light source infrared LEDs, a position-sensitive detector for converting the location of the optical input into an electric signal, and the amplifier circuit were integrated in the sensor body. </p><p>The scattered light arising at the point of contact is transmitted to the detector through a bundle of optical fibers. By processing the detectors electric signal by computer, it is possible to determine the contact location on the sensors surface in 1.5 milliseconds. Through further miniaturization, a fingertip diameter of 20 millimeters has been achieved in the latest version of the tactile sensor. It is currently planned to install this tactile sensor in a robotic hand with the aim of improving its dexterity. </p><p>Instructor (Oussama Khatib):Okay. A cool idea, right? Because now youre taking this information, and taking it into a visual image, and transmitting the image, and, in fact, this was done a long time ago. I believe the emperor of Japan was visiting that laboratory, and he saw this, and he was quite impressed. </p><p>Before starting the lecture, just wanted to remind you that we are going to have two review sessions on Tuesday and Wednesday next week, and we will, again, sign up for two groups. I hope we will have a balance between those who are coming on Tuesday and Wednesday. We will do the signing up next Monday, so those who are not here today, be sure to come on Monday to sign up, all right? </p><p>Okay. Last lecture we discussed the controlled structure. We were talking still about one degree of freedom, and we are going to pursue that discussion with one degree of freedom. So we are looking at the dynamic model of a mass moving at some acceleration, X double dot, and controlled by a force, F. So the control of this robot is done through this proportional derivative controller involving minus KP, X desired and minus KV, X dot. So the KP is your position gain, and the KV is your velocity gain. </p><p>Now, if we take this blue controller and move it to the left, the closed loop behavior is going to be written as this second order equation, and in this equation, we can see that we have, sort of, mass, string, damper system whose rest position is at the desired XD position. So KV is your velocity gain, and KP is the position gain. </p><p>Now, if we rewrite this equation by dividing it by M, we are going to be able to see what closed loop frequency we have and what damping ratio we have, and every time, the lecture time, this finishes. So what is your closed loop frequency? KP is equal to 10, and the mass is equal to 1; what is the closed loop frequency? </p></li><li><p>Student:Square root of 10. </p><p>Instructor (Oussama Khatib):Square root of 10, and what is the damping ratio? A little bit more complicated, but we can rewrite this same equation in this form, 2 zeta omega and omega square where omega is your closed loop frequency, and where zeta is this coefficient, KV divided by 2 square root of KM, and omega is simply the closed loop frequency square root of KP divided by M. </p><p>So you remember this, but now the difference with before, before we had natural frequency, so we were talking about natural frequency and natural damping ratio. Now, this is your gain, and you are closing the loops, so this is your control gain; its the closed loop damping ratio and the closed loop frequency, okay? So the only difference is instead of a natural system with spring and damper, now we are artificially creating a frequency through this closed loop, or we are creating this damping ratio through KV. </p><p>So, basically, this is what you are going to try to do, you are going to take your robot; you are going to find those gains, KP and KV, and try to control the robot with those gains. So, again, thinking about KP and KV, KV is affecting zeta, right? And KP is also affecting your omega. Now, when you are going to control your robot, what is the objective; what are you going to try to do? Lets think about it. Youre trying to go somewhere, right, or you are trying to track a trajectory. So what do you want to achieve with those, I mean, here is your behavior; what would be good to achieve here? Yes. </p><p>Student:It could see in critical damping. </p><p>Instructor (Oussama Khatib):So we want to have a critically damped system most of the time, so we will reach those goal positions as quickly as possible without oscillation. So KV would be selected to achieve that value, and for that critically damped system, what is the value of zeta; anyone remembers? It was only two days ago. Zeta is equal to for critically damped systems, zeta is equal to unity, 1. When zeta is equal to 1, that is when KV is equal to 2 square root of KPM, you have critically damped system. </p><p>So, basically, if you know your KP, if you already selected your KP, and if you want critically damped system, then immediately you can compute KV from M and KP, right, for that value, for zeta. So, basically, you are trying to set zeta. What about omega? So now, we need to set KP in order to compute zeta, and how do we set omega? Someone? No idea? So you have your robot, you go and you want to control, lets say, Joint 3. We can do it if you want. Wheres my glasses? Heres the simulator. Oh, that doesnt have an F factor. Lets take this one. </p><p>So, here are your gains, and right now, if we ask the robot to so, the robot is floating, and if we ask the robot to go its zero position, its going to just move, and its moving with a KP equal 400 and KV equal 40. These are the gain we set for the robot, but, in fact, this is controlled also with dynamics. So we will get to this a little later, but if we want to see the control without dynamics, we take this, probably, non-dynamic joint control, so this one. </p></li><li><p>So lets float it a little bit. Actually, I can exert a little force outside and see if it can move; its really solid. Well, okay, wont move it too much. So lets reduce the gain here. So this springiness KP is 40. So see, now if I apply a force that is a deflection, right? And when Im going to release, its going to go there, oscillate a little bit, tiny bit, not too much. In fact, this has a lot of friction, natural friction. If we remove the friction and do the same thing, it will probably oscillate more hm, not enough. Okay. Wow, still there is friction nope. So lets put a little bit, minus how much? Minus two, this is -20; I think it will go unstable. Wow. So we see that your gain cannot be negative. It will can you stop? Okay. We need some friction, otherwise it will not stop. </p><p>So, in fact, you can see there is a lot of coupling. I moved just one joint, and everything else is moving. Lets make this gain bigger. This is Joint 1, so if I pull on Joint 2, and the release look at Joint 3; what is happening? So there is an inertial coupling coming from Joint 2 on Joint 3. Just by moving Joint 2, you are affecting Joint 3. You can see, again, Joint 2, release, and Joint 3 is moving. So in order to avoid that disturbance coming from the dynamic, what should we do with KP? Make it smaller or bigger? Youre not sure. Should we try it? </p><p>So lets make it bigger; how big? 400? Okay, 400. Now we realize with 400, this is not damped enough because we need to compute this to make it a little bigger, so lets make it 20. Okay. So now, what do you expect; the disturbance will increase or will be reduced when I am going to release? More disturbance or less? Heath, less? </p><p>Student:Less. </p><p>Instructor (Oussama Khatib):Who agrees with less? Okay, and who disagrees with less? Everyone else, okay. So this is less? Yeah, it is less, actually. Youre removing little faster, and you are still oscillating, and oscillation is because we dont have enough damping here. So if we increase the damping, it will oscillate less, and if we increase the gain do you see what is happening now? Its going very quickly to its position. </p><p>So, in fact, the coupling this is the degree you look at the 90 degree between Joint Link 2 and Link 3. It is maintained, almost. In fact, if I increase Joint 2 as well, it will be hard to move it. So what is happening now with the response; do you see the response when we went to 1600? Faster or slower? Hm? Slower? </p><p>Student:No. </p><p>Instructor (Oussama Khatib):Faster. So the dynamic response of the closed loop is faster with higher gain. Well then, should we increase it, like, keep increasing? I dont know. We can try. </p><p>Student:But theres a limit at some point. </p><p>Instructor (Oussama Khatib):So what is the limit? So lets make it 3,000. Now, Joint 3 is locked; its not moving anymore. Should we make it more? Okay. So whats going to </p></li><li><p>happen? Its not moving anymore. Now, the problem if this was a real robot, would 30,000 work? Why? </p><p>Student:Your motors gonna saturate at some level in </p><p>Instructor (Oussama Khatib):Well, suppose you have big motors. Yeah, saturation of the motors is one thing, but suppose you have really big motors; its not a limitation. </p><p>Student:Wouldnt you have some sort of air drift? </p><p>Instructor (Oussama Khatib):Well, well discuss it a little later, but, essentially, what is going to happen is that remember, inside the structure you have motors, you have transmissions, you have gears, and all of these are going to move, and they have flexibility in the structure. This flexibility makes it that you start to excite those mode of the flexible system, and as you start moving, the motors start to vibrate, and if you have flexibility in the structure, the structures start to vibrate, and when you hit those frequencies of vibration, the system will just go unstable. </p><p>So our KP, this KP that we want oh, we closed it. Just one second, lets go back there. So this KP we have here, this KP cannot go too high. We want it as high as possible to increase what? What it does when KP is high? Disturbance reduction because errors are coming dynamic coupling coming from other links will be rejected; its stiffer. However, a KP cannot go too high because KP is deciding the closed loop frequency, and this closed frequency can go as high as those end-modeled flexibilities. Actually, we cannot even come close to them; we have to stay away from them. So omega cannot be too high, which means KP has a limit, but we want to achieve the highest KP. </p><p>So what is the relationship between KP, KV, and those performance? So from those two equations, we can write KP is M omega square, and KV is M to zeta omega, right? Just to rewriting these two equations and computing KV and KP. So when we are controlling a system, we are going to set what? Were going to set, really, the dynamics of the system, which means we need to set zeta and omega. So we set zeta and omega, and we can compute our KP and KV. Most of the time, zeta is equal to one. So KV is M to omega, and so all what is left is to set omega. So for 400, omega is equal to what? In the case of the robot in this simulation, we have 400 KP. So omega is equal to? Come on. </p><p>Student:[Off mic]. </p><p>Instructor (Oussama Khatib):Square root </p><p>Student:[Off mic]. </p><p>Instructor (Oussama Khatib):Divided by well, M is equal to 1, lets say, in that case. Its 20. Its 20 multiply what is the frequency, the real frequency? </p><p>Student:[Off mic]. </p></li><li><p>Instructor (Oussama Khatib):Omega divided by 2 Pi, right. So what is your frequency about lets say divide by 6, 20 divided by 6. So its very low, 3-4 hertz. In fact, if youre lucky, you can go, well, to 10 hertz. I mean, this would be great. So when we go to 1600, this is really nice, 40 divided by 6. </p><p>Well, in practice, you start with very low gains, and you start turning your gains up, up, up, up, and suddenly you are going to hit that, noise..</p></li></ul>