 Let us quickly summarize what we did in the last class. We had looked at what is called as the state space approach. The importance of state space approach today comes from the fact that it is extensively used in the control algorithms in understanding automotive controls and so on. So, we have to digress a bit and look at state space. It is also very easy to understand how for example, a vehicle behaves for a particular manner. We will see that in a minute. Let us finish this. We will see that how this can be used. Now, what we have looked at this particular case is a simple bicycle model where we had the 2 degrees of freedom called as V and R. It can be replaced of course, by say for example, beta. You can replace beta and R and so on. You can also extend this. It is not necessary that you have to have only 2 degrees of freedom. You can extend it. You can go to 4 degrees of freedom or you can go even up to 11 degrees of freedom, where you would introduce the tire characteristics and so on. So, in other words, it is difficult in a class to derive all those things. There is a reason why we have taken 2 degrees of freedom. We will indicate how it looks like, maybe 2 classes from now. How it would look like if there are say 4 degrees of freedom? Let us take a, I will just give you the final expression. You can always derive it is not very difficult. So, what we are now doing is that whatever be the number of degrees of freedom of the fundamental model which results in the what we call as the state space equation from that. The form of that equation can be written like this. So, the form of the equation can be written like that. So, the general approach that we have now can be extended to whatever be the degrees of freedom. That is the whole idea. So, the state variables that is x, the vector x can be any of those things. We will see that a bit later. So, the approach is general. That is why we are putting this down. But in a class, if you want to work out, we will stick to 2 dimensions or 2 degrees of freedom and that we will see it later. So, that is the fundamental equation and that is the output. All of you know, we went through this x t, we determined or we derived this. Remember that we had Laplace transform and so on. So, we have 2 terms ultimately for x, you know the x consists of 2 terms. One is what is called as the 0 input term. In other words, this is purely because of the initial condition, 0 input term and this is the 0 state term. In other words, this is due to the input u. So, the response consists of the initial as well as the input conditions. Now, all of you know, I need not repeat it, may be for the sake of some of you. Remember that we had in the one, I mean if there is a, for the differential equation, we had a solution e power At or alpha t rather At rather. I think I put A, right? e power At and that if A happens to be negative, then the solution does not become unbounded and so, you had that stability criteria in your earlier classes that the real part of the Eigen value should be in the left of and so on and so forth. We will come to that left of the z plane and so on. You know all those things, just to remind you that they all can be now factored into these equations, right? That is the state variables and then we ultimately looked at the linear time invariant system, which is our system and identified that e power phi t is equal to phi of t is equal to e power At, right? And wrote down that and determined also what we called as the state transition. This is the state transition matrix as I had indicated here is called as state transition matrix because of the fact that how it is participating in the expressions and we remember that we said phi of t is the inverse Laplace transform of psi of s, remember that that is what we did and remember that in which case that expression now boils down to this expression. This is all we did before and that y of t can be written as this standard form in the state space form and so that x of t is this you know that is specifically to this from which I can write down y of t, okay? y of t is C e power At x0 plus 0 t C e power At minus tau B u tau, right? Okay. Now one of the interests in any of the dynamic system is the impulse response, okay? Why is that we are interested in impulse response? That is one thing but more important also is that we can find out the transfer function, frequency response function, okay because the Laplace transform of an impulse is what? Unity, okay? So directly we can get the transfer function from the you know if the impulse is equal to if the Laplace transform of the impulse is equal to 1 then you can easily find out the transfer function and so on, okay? Let us not repeat that you know those things already, okay? Now I want to find out what is the impulse response of the system, right? So how do I find out the impulse response of the system? So what is the input now? Input is the impulse, okay? I can write down that and I am not going to derive it completely. I am just going to write down only the shifting property which you all of you know phi of t, delta of t minus tau is equal to phi of t. This is because of the what is called as the shifting property, okay? Now substitute that and then into this expression and ultimately you can write down C e power A t minus tau B plus D delta t minus tau. For a minute we will remove this D, okay? D is not usually is not there anyway if you want to retain it, you can retain it. How did I get this G from this is nothing but Y, okay? So from Y I determined that expression, right? And then I write down that as G. Now by a change in the variable I can write down change in the variable I can find I can write down, okay? So that is the impulse response of the system, okay, right? We will come back to this. We will come back to this because we will let us not get lost in further equations on this. Just keep that in mind, okay? We will again go back to hardcore vehicle dynamics, okay? And then expand this later, right? We will develop a simple technique to understand. Let us go back to physics, okay? Too much of maths, too much of this and you will lose track of what we have been doing, right? Okay? Now let us look at important quantities. Let us look at the physics of the dynamics and then keep that in mind. I will come back to that equation a bit later, right? Okay? Get back to the expressions that we had used before, okay? So let me write down that expression. I am going back to the A, okay? Remember that we had V dot r dot x dot is equal to A x, you know? I am going back to that expression with a Laplace, you know after Laplace transform. In other words, let me get back to this expression, okay? Clear? What did I do? I just took a Laplace transform, recognized that x dot of t Laplace transform is s into x of s, okay? Substituted it in that expression, I have just rearranged it, right? Nothing very difficult. Now I am going to slightly digress from here and I am going to get the steady state gains, okay? And what is steady state gains and why am I interested in this? So let us get back to the physics of that expressions. Now we are looking at maneuverability, okay? Lateral dynamics. Suppose you are driving a car, okay? There are two situations, there may be others, you know, two situations where the maneuverability of a car becomes important. For example, you are doing a lane change, okay? You are doing a lane change. Say for example, you are here, you are here and there is a car before. Then I have to do overtake the car, okay? I am doing what is called as the double lane change, okay? This is one situation where you want the car to respond to your input, the driver's input, okay? So that you will have no issues in overtaking the car. The other one, the other one is that when you take say, for example, a sharp turn, you take a sharp turn, okay? Why am I putting these two as if they are different cases? They are different cases where you as a driver would like the response to be, the response of the vehicle to be to have two different aspects. Your input is the same, you are going to give the steering input, okay? Your input is the same but you are expecting the vehicle, vehicle to respond in a slightly different fashion or from a dynamics point of view, you are going to look at two different things. Let us take this case, it is easier first, then come to that case. What are you trying to do here? You are reorienting the vehicle, reorienting the vehicle as you do that whenever, okay? You are reorienting the vehicle which means that what is important here? It is the yaw, okay? The yaw performance of the vehicle to your input as far as the yaw is concerned becomes important, right? So, that is what the yawing of the vehicle or what becomes important is the rate at which it goes or the yaw of the vehicle, okay? For your delta input, what is delta steering input becomes important. So, in other words, here you expect the yaw to be reoriented, oh sorry, raw to help, yaw to help the vehicle to get reoriented in order to take this path, okay? You as a driver, you are looking at that, not that lateral acceleration is completely removed. Of course, lateral acceleration is also important but response for you, the response is, am I taking that turn correctly? Am I, is the vehicle following that? That is what becomes important. Here, more than the yaw, it is the lateral acceleration becomes important. Here, it is the lateral acceleration because you do not want to go off the path because this may be done at a higher speeds and so the lateral acceleration as you take that curve, okay, that becomes important. Yaw is not that important. It is not that they are not important, okay? Both of them are important but the relative importance will be different in each of these cases. So, in other words, what is that you are interested in? Given a delta, in both cases you are giving delta, you are giving only the steering input. Given the delta on the front, okay, what is the response of the vehicle as far as the lateral acceleration is concerned? Either you can write it as u squared by Rg or u squared by R does not matter, okay, u squared by Rg or R, this is u squared by Rg or u squared by R, okay? The response of the lateral acceleration and what is this response of this to yaw, right? So, these are the two things that become very important for us. So, that is R, right? I mean, what is the response to this? Now, how am I going to determine that from this expression? In other words, we call this in our control system language as the gain, okay. What is this? That is the lateral acceleration gain, okay and what is this? These are the two things that become important for us, okay from the vehicle response. You can do exactly the same thing with these expressions. This is not a problem but I will make it slightly simple so that we will not lose the physics. I will come back to a more elaborate state space a bit later, right? Why do I keep writing only delta F? You know, does it mean that delta R, okay, is also there? Yes, I had already told you that delta R can be for a very long vehicle but interestingly there is another condition, your own vehicle, I mean, what cars you drive, they do not have a rear, you know, staring at the rear but that becomes important when you back up, when you are reversing your car back up then actually what are you doing? You go, okay, in the U direction is now in, you know, in the negative of you, that is another words you go, you back up and then how do you steer? The rear steer, okay. So this is, in other words, this is a condition which is also common in everyday use of a vehicle, okay. We always find it difficult to back up and it is very easy to drive in the front. Not that it is difficult to see even if you have a camera before you, sometimes it is difficult to back up. Why is it? Keep that in mind. We would like to answer that later. Right now we are looking at only delta R, I am sorry, delta F, okay, input in the front. We will then see what happens to delta R, clear? Okay, how do I now determine this transfer function? How do I determine the transfer function or in other words, this gain, the respect, yes and so on. What I am going to first do is to find out what is V by delta, lateral velocity. Because once I know that, I can always find out A later, V dot plus R U, V dot is SV and so on. So it is not going to be very difficult. So I am going to find out what is V comma delta and R comma delta. This is what I am going to find out, okay. So how do I do that? Simple, delta F. I am going to use the Kramer's rule. I am going to put, okay. Let me say that, let me call this as delta, okay. This whole thing is delta, right, this matrix and if I want the first expression V by delta F, I put this, this guide into the first column, okay and use that as the numerator and use the determinant of delta as the denominator. This is what we are going to do, okay. So if I want now R by delta, then I am going to put this here for that column, retain this column and determinant of that, I will use that as the numerator and determinant of delta, I will use that as the denominator, okay. So that is what we are going to do. Later to look at a very interesting physics, we will add, once we know this, we will add delta R, we will do that separately and then we will use this technique in order to find out what happens to V by delta R and R by delta R, that we will do it a bit later, right. Any questions? Do that. So V by delta F is equal to substitute that, multiply it, okay. I am not going to do all those multiplication. You do that so that I will write down the final expression C alpha F, yes. That is the first term, okay. So what do I do? Obviously I C alpha F, AC alpha F, then MU plus. So that is the determinant of that is what is my numerator, okay. Determinant of this, find out. I hope my signs are correct. Just check this whole thing divided by the determinant of delta or let me call that as determinant of delta, the delta matrix, that is the, right, okay. R by delta F, I did the same thing, put this here and I am going to write down the final expression, check this as well. If there is any questions, I will answer that because of lack of time, I am not going to substitute it, derive it and so on, that you can do it very easily divided by determinant of delta, okay. Now if I want the lateral acceleration, then V dot plus R U divided by, remember that is what we had in our expression is equal to delta divided by delta and V dot is nothing but SV, right and plus R U, R I know U. So combining these two, I will get V dot plus R U, that is the lateral acceleration gain, okay. I can find out that from these two expressions, can I write down, what is that expression? Look at this SV, so I Z C alpha F S squared, okay minus M into A, what is A? A is not acceleration, remember that A is the distance from the C G to the front axle, okay. M A C alpha F by U into S, right plus A plus B B C alpha F C alpha R into S divided by U, here R into U, right. So you would ultimately get an expression of this form that I Z C alpha F S squared plus A plus B into B C alpha F C alpha R S divided by U plus A plus B C alpha F C alpha R, the whole thing divided by determinant of delta is, that is going to be a huge expression. Anyway, this is not a very difficult, nothing very difficult to get, right. That is the determinant, this is the numerator and that is the denominator. For R by U, just substitute here for this, substitute that. Just check that, let us check this, that is why I said put that here, okay C alpha F, so I Z S plus A squared C alpha F plus B squared C alpha R by U, okay. Substitute it and check, what is the expression, okay. So now, multiplying this I Z, so the first term is fine, first term is fine. So the second term is A squared, so let us do that. The first term is I Z into S into C alpha F, right. The second term plus A squared C alpha F squared plus B squared, right. C alpha F C alpha R, right, divided by U, that is the second term, minus, what is the first term, A C alpha F into M A C alpha F into U, right, minus A squared C squared alpha F, so plus A B C alpha F C alpha R divided by U, is that correct, right. So the first term that is correct, okay. Second term should be U, not divided by U, right, that is what, right, okay, sorry, that is the multiplied by U, then this term goes off, this term goes off, so you have that there, right, okay, good, sorry, so that is good, all of you are awake and kicking, good, great. So now you add this, I think you, now V dot S, so I Z C alpha F S squared, yeah, that is there, okay. Now this becomes minus A S C alpha F U, okay, right, because here I have same thing, M A C alpha F S into U R U, okay, there will be a U here, so this guy and that guy will go off, right, so I will have absolutely, so that is an important thing, so now you see that this term is multiplied by S and that is the term I have here and this term is multiplied by U and that is the term I have here, right, correct, great. Now let us look at this, thanks and a small, okay, clear, now you are clear with the direction as well. Now let us quickly pick up this and this before we write what is steady state gain, what is steady state gain, when do we say what is steady state gain, this is nothing but the transfer function, so when do you call this as steady state gain, when S is equal to 0, that is all, okay, so that means that R by delta F, this term will go off, can you tell me what is the steady state gain, A plus B C alpha F C alpha R divided by U divided by, right, the last term here, L squared C alpha F C alpha R 1 plus K U squared by U squared and you can simplify that later, that is at S is equal to 0 and similarly you can write down A y steady state gain for A, A also, amen for the lateral acceleration as well but before we go further, let us look at these two expressions and what do you pick up from these two expressions, it is a very interesting thing that is happening here, look at the numerator, it is first order polynomial, that is in other words you have only S, here this is second order, okay, what about this numerator has a second order and the denominator has a second order, okay. Now all of you know that if I want to convert this into a frequency response function, what do I do? S is equal to j omega, fantastic, so I will substitute S is equal to j omega in order to convert it into a frequency response function, right, so what do you pick up from that? Say for example, when the frequencies are high, when the frequency is high, okay, it is a frequency response function, the frequencies are high as the frequency goes higher and higher, what happens to this response and what happens to this response, this would go to 0, yes, so the frequency response at high frequencies, okay, this would go to 0, which means that there would not be any yaw velocity for delta f, what about this, this would become a constant because the first two terms would go off, okay, that would become a constant, okay, so at very high frequencies this goes to 0 and this goes to a constant, so if you want to do any of the, you know, sensors to be placed in order that you want to control, okay or you want to look at the automotive control, okay, then what is the sensor that you would use or what would you pick up? Would you pick up yaw or would you pick up lateral acceleration? You would pick up lateral acceleration because lateral acceleration grain is still, you know, it is a constant, you would pick up that, okay, so write down for both this as well as for, so let me simplify this, that is equal to a plus b is l, so u divided by l into 1 plus k u square, remember that k is the understeer gradient, sometimes called as understeer coefficient, okay, we will use this as understeer gradient or understeer coefficient, why do we call this as understeer gradient? What is this gradient about? Why is that we call this as understeer gradient? Remember that graph which we talked about in terms of delta versus u squared by r, you remember? Go back and check that graph, okay, remember that I had delta plus u squared by r, delta versus u squared by r, so in other words, delta is equal to k into u squared by r plus l by r, you know, that was the expression that we got, right, remember that we had drawn it, u squared by r versus delta, the graph was like this, remember, like that, the graph was like this, right, so what is this gradient? In other words, what is the slope of this? That is the k, right, so because of the fact that it is the gradient in the graph between delta versus the lateral acceleration, we call this as this understeer gradient. People sometimes call this as understeer coefficient, it does not matter, but remember that that is the, you know, word that is used, right, right, okay, so let us write down the steady state gain for a1, so I will call this as lateral acceleration divided by delta f that s is equal to 0 is equal to u squared by l into 1 plus k u squared, substitute it and you can get that, right, clear, okay. One of the things that you learnt in your earlier classes is the importance of this, we will get back to the equations in a minute, but the importance of this denominator, right, remember that you had, this is the, what did you call this as, characteristic polynomial or characteristic equation, remember that you had equated this to 0 and remember that you had what are called as poles, okay, so the numerator and the denominator have interesting properties, okay and remember that when you have equated this, you got poles and what about the numerator? When the numerator is equated to 0, what is that you got? Zeroes, so poles and zeroes are obtained by equating the denominator and the numerator to zeros, right. Remember also these are some, I mean I am just recollecting what all you did in your earlier class on controls. Remember also that poles had an important role to play on the stability of the system, right. So, you, what you did was to plot in the s-plane, okay and you are always worried that the poles should line to the left of the system of the, I mean the left half plane, okay of the s-plane, right and that is fine, not very difficult to understand because the poles actually go to determine the solution or the response and if it is negative, e power minus say 2t would die down and hence we understand that that is important. So, in other words, we will come back and in other words what we did was to look at the stability of the system through the poles, okay of this system. We never bothered about zeros actually, okay. In our earlier classes, we never bothered whether to see whether zeros were to the left of this, of this plane or to the right of the, of course it can be, right, left and right, left half plane or right half plane. That is going to be very interesting, okay when we look at instead of delta f and we look at delta r, okay. So, in other words, when you are backing, you are going to find a very interesting twist to the zeros. What is it? After a minute or after a small derivation will come back. So, to summarize, this is very important for us to determine the poles, okay. The response of which of the system is determined by these poles, the position of the poles and that of zeros, okay and what happens if this zero falls on the right hand side of the plane, can it fall? What happens? We will see that now, okay. Any questions so far? Okay. Now, we are going to look at a response of the system. We will generalize it later for, this is for the bicycle model. We will generalize it later for a much larger system by expanding our the, what we did so far. I am going to write down these equations which you already know. That is the one problem with this course is that you will keep on writing down the equations which we derived before and I hope you are, we are in track if you have any doubts stop me, okay. We will review this aspect. What I am going to do is to pick up what all I did so far in the state space, okay. So, just to summarize, just to go back again. So, if I want the frequency response function for any of these things, no issues at all. I can get this from these expressions, okay. It is just substituting s is equal to j omega, okay. So, these two responses are important as we said, okay. What is, if I have to take a double lane change, then the expression which I would be looking for, the response I would be looking for is this. If I have to do, say for example a cornering, then the gain is this, okay. So, that is the, that is the expression which I would be looking for, right, okay. Now, ultimately our aim, what am I going to do? Let me summarize that and maybe we will start the derivation in next class because I am going to go through large derivation. Ultimately my aim is to understand handling, okay and see whether I can put down a quantitative measure for handling, okay. Yes, this is fine. You understand the physics. You understand that yaw becomes important. You understand that lateral acceleration becomes important and so on. But ultimately I want to understand handling in a very quantitative fashion. What does it mean? It means that if I have two or more cars, okay, if I come and give you, say three cars and tell you, tell me which handling, which of these cars has good handling characteristics. Suppose I ask an question like that. Tell me which has good handling characteristics, okay. Now, you can drive the car, of course and then tell me your feel, okay. Looks like this car is good. You know, in other words, you can give what is called as a subjective evaluation. You can go to and give me a subjective evaluation. This is just by driving the car. Drive the car, take a turn, you take a double lane change. There are ISO standards for this. You do a slalom test, you do a pulse test, you do all those things and say, okay, I feel this car A is good then, is better than car B and so on. There is other thing which is called as an objective evaluation of the vehicle. Okay. Now, what is objective evaluation? I have these things. Yes, of course, these equations. Okay. What does it really mean, okay, when I have two or three cars? How am I going to look at the response through these equations and tell, okay, quantitatively car A is better than car B, better than car C and so on. Okay. Why are we interested in an objective evaluation? Because objective evaluation very clearly tells you what is the characteristics of the car which makes it good or bad. Subjective evaluation by the driver would only tell you whether this car is good or that car is good in a language which are not necessarily very technical. Of course, every car company relies on subjective evaluation but it is very important to now bridge this gap and marry these two. Subjective and objective evaluation. So, we are now looking at handling from a very quantitative perspective with all the equations which we have derived, okay, and put down the rules and say that, look at these four, five, six quantities. I have a number of them. I have a yaw, I have a yaw rate, I have some yaw rate gain, I have lateral acceleration, I have beta, V by U, the vehicle slip angle and so on. So many things that now I know. What is important? How am I going to now combine them and tell that vehicle A is better than vehicle B? So, in other words, how am I going to convert these things into characteristics of the vehicle? That is what we will see and we will pick up one of the models and explain that model, okay, and understand that model, okay, and how we can use that for the vehicle evaluation. We will do that in the next class starting with my derivation. Clear? Okay. Please revise the state space because I am going to use the state space again, clear? Okay.