 Let us review quickly what you know on the state space technique so that we will go forward applying this. Of course, we had already discussed a few of the points. We will just put down all the equations just to revise what all you have done in the previous classes on some controls. So, this all of you know this is the response y of t and the impulse response function and this is the convolution term. You know the convolution part of the response which when we take the Laplace transform of that equation you get the multiplication of this convolution term or convolution integral and we also saw that the Laplace transform of e power at is Si minus a inverse right and that the g of s which is the impulse you know transfer function is given by this equation. Go back and look at it. d is usually equal to 0 in our cases. So, that can go off and very standard equation comes from remember x dot is equal to ax plus bu y of t is equal to cx plus du from that we get this. And the poles of the system are determined from this determinant, determinant equal to 0 and you know the importance of poles and once let us call p 1 p 2 be the poles of the system and if p 1 p 2 are the poles of the system then g of s is written say for example in partial fraction form and that if you now take the inverse Laplace transform you get g of t which is given by this function. In other words the poles play a role which are loosely called as the Eigen values of this things poles or Eigen values. One of the questions which people have especially when they come from a vibration background is that of the connection between the natural frequency and these poles. There is always a confusion as to what we mean by natural frequency and damping. In other words the poles of the system are expressed in a slightly different fashion okay as omega n and zeta for example omega is the natural frequency and zeta is the damping ratio. So, the relationship between say for example if the poles of course you know that it has to be a complex conjugate okay or if or it can be only real in which case you are looking at the time for relaxation 0.66 and so on you know that is given by this time and more importantly if you want to pick the relationship between the poles the natural frequency and damping then the relationship is given by this form okay. So, it is people usually say that natural frequency is the Eigen value or poles and so on they are not strictly correct yes one comes from the other but please note that if these are the poles okay the complex conjugate poles then this is what is the natural frequency okay. Now you understand the natural frequency and the poles you know what are important okay. Further than this about the effect of poles you all know it and that you know that in the S plane if this is the left of S plane then the poles are situated in the left of S planes S plane the system is stable basically because you have this e power term here and hence when it is negative okay the e power minus term minus into t would now start decreasing with time and hence okay the system becomes stable okay. Electrical engineers have very interesting analogy or very interesting way of interpretation not analogy but interesting way of interpretation how they interpret this. So, this is the j omega okay. So, you would notice that if I want to look at the frequency response what is that I do I replace S by j omega right. In other words when I want to plot the frequency response then I what do I do I plot omega here okay and the response in the y direction or in the y axis. So, you would notice that actually as I go along omega I go up that axis okay I at various places okay I go up this axis as I go up the axis if you look at what actually I calculate okay I calculate the distances between the poles okay between that point and the poles that is what is meant by the denominator actually that is the distance. You would notice that as I go along okay above this and when I come to a minimum distance point a minimum distance point okay assuming that these distances are large to the other pole to the other pole the distances are large okay since it is in the denominator okay as I move along omega my response actually keeps increasing because the distance keeps decreasing okay. So, when I come to this point because as I go remember as I go along omega actually I am going above in the y axis. So, when the distance is least I get the maximum response okay in other words that is why you have that remember that is the omega that you would get the maximum response. So, then as I move away the response now starts dropping okay. So, when there is no real term which means that your poles lie on the imaginary axis what would happen that it would go to infinity and so you would have a graph which goes like that and comes down. So, this is another interpretation okay especially used in filter design where it is very simple to understand okay the filters which is low pass filter or high pass filter or band pass filter for example okay those things can be band pass filter can be understood very easily by these distances okay which are in the denominator right and which one by the denominator one by the distance is small then the response is large okay after all you are looking at the response in the left hand side okay. So, that is a neat interpretation for the frequency response function okay. In the last class any questions okay please revise once the state space formulation if you have any doubts okay. In the last class we were looking at remember the transfer functions right we said that the two important transfer functions that we had looked at is for lateral acceleration okay lateral acceleration we in fact what we did was to find out the lateral acceleration versus delta front okay as well as what is the other one we did r by delta. So, I can write this as Ay just for simplicity Ay by delta and r by delta remember that we had only the front delta f and so on right. Now we also if you remember right we also wrote down the steady state gain okay right we wrote down the steady state gain the steady state gain at s is equal to 0 is u squared divided by L into 1 plus k u squared and that is equal to u divided by L into 1 plus k u squared right this is what we wrote down right. Now what I am going to do is to apply whatever we have done here okay to whatever we have done so far in other words I am interested to find out the folds of the system and then determine the omega n and the damping characteristics of the system this is what I am going to do now okay why are we doing this remember in the last class we said that we need to understand this whole concept of the response from an objective perspective okay from an objective perspective and hence we need to understand how actually the system behaves okay and can be written in terms of omega n and delta okay I am going to write down now a series of equations very simple algebra okay so I am not going to derive that step by step so that is going to take a lot of time so I am going to write down all these equations because interpretation becomes important so I am going to do that later okay one of the things I promised that we have to look at also what are called as the zeros right what are zeros the numerator okay the same what you did with the denominator you do that with the numerator okay we have not looked at the zeros yet we will have a very interesting thing what happens to the zeros that we will do it okay after we finish this derivation we are going to follow for objective there are a number of objective metrics as they are called we are going to follow this paper by Mimuro Mimuro et al SAE 910 sorry 901 734 this is not the only way of looking at objective evaluation but it looks you know quite simple quite appealing and hence I am taking this paper right so the first step let me write down what all you already have it so I do not have it in memory so I am going to write down what is A matrix for our problem if there is any doubt go back and refer this are all there in your notes right so this is this is A actually if you look at literature there is a lot of confusion on this C alpha f and alpha r I was looking at a thesis the other day there is a lot of confusion people have put C alpha f and alpha r to be negative okay in the the signs are given in such a fashion that C alpha f and C alpha r are given as negative in which case the whole understeer oversteer gradient definitions change okay so this looks like an accepted practice but I would like you to I would like to warn you that when you read paper or a book be clear okay like how is this definition or given you know what is this definition we are taken minus alpha f so that ultimately that equation becomes you know plus C alpha f okay sorry we taken minus alpha f and so C alpha f we kept it as positive okay so that we get the force centripetal force accordingly but there are people who have written this alpha f to be positive and written C alpha f to be negative okay in which case their k definition would change and so on just a bit of a bit of warning so when you read a book ultimately everything would boil down to the same thing but but then the same interpretations but then you have to be careful on the signs because when you do a problem or you want to understand it you know things would be different you have to be careful on that right. So the next step is S i minus a put that here C alpha f plus C alpha r divided by m u a c alpha f minus b c alpha r divided by m u plus u a c alpha f minus b c alpha r divided by i z u s plus a squared c alpha f plus b squared c alpha r divided by i z okay so that makes my S i minus a so the determinant of S i minus a is equal to 0 is what I have to find out okay that is my next step I am going to do that. Modern vehicle dynamics follows the language of controls very closely in fact if you look at older books okay the approach would be very different state space approach is usually not followed in older vehicle dynamics books but today because of the importance of automotive controls most people have shifted to the language of controls and hence we are talking the language of state space okay makes also very this whole subject to be very elegant because you are just applying what you studied in controls to vehicle dynamics. So determinant of S i minus a is my next step do that follow me and I hope I am correct S squared plus C alpha f plus C alpha r one step I am just removing it m u plus a squared c alpha f plus b squared c alpha r divided by i z u s plus c alpha f plus c alpha r divided by m u into a squared c alpha f plus b squared c alpha r divided by i z u minus this is the last term minus a c alpha f minus b c alpha r divided by i z m u u into a c alpha f minus b c alpha r divided by i z u I hope it is correct just check this I think I am correct on that yeah not very difficult to check just check that which one this will become s plus so s plus c alpha f c alpha r s squared correct then a c alpha f alpha r by m u right plus a squared c alpha f plus i z u yeah yes right plus c alpha f c alpha r that is the last term is from there u into plus so minus is taken out so that would be minus is outside this one now what do what say which which one this term clear okay the problem is there are so many terms you have to be careful this is nothing nothing very difficult about it yes right most of the times m is usually accompanied by u squared you are right okay that should be the case now let us concentrate on this last term okay let me let me simplify this last term okay that is what I am going to do simply simplify it for me simplify this for me so this would be just follow me very good excellent you know just follow me whether all my steps are right so a squared c alpha f squared plus I know the last step is correct but let us see that all the steps are correct plus b squared c alpha f c alpha r divided by i m u squared minus a squared c alpha f squared plus b squared c alpha r squared minus 2ab c alpha f c alpha r minus m u squared ac alpha f plus m u squared bc alpha r whole thing divided by i m u squared okay simplify it beauty of it is ultimately this is the third term the last term okay ultimately this happens to be l squared c alpha f c alpha r into 1 plus k u squared by i m u squared right check clear okay you have to bear with me till we finish all this then we take the second term and I am going to simplify it remember what is my goal my goal is to find out ultimately omega n and zeta I want to find out the poles from poles I can find out or I am going to develop a much simpler technique in order to find out the natural frequency and the damping ratio okay right. So, the second term is that is the term so let me take m i z u squared is common m right sorry m i z u is common i z u is common okay so the multiply the first term by i z so that I get i z into c alpha f plus c alpha r that is the first term plus the second term is m into a squared c alpha f plus b squared c alpha r okay that is the second term let me rearrange them so that I will write that as i z plus m a squared c alpha f plus i z plus m b squared into c alpha r divided by m i z into u right nothing difficult just rearranging the term. So, now I am going to put these two terms okay simplified terms into my determinant okay and write this down as a squared plus this whole term i z plus m a squared c alpha f plus i z m b squared c alpha r divided by m i z u s plus the last term l squared c alpha f c alpha r 1 plus k u squared divided by i m u squared is equal to 0 okay. Now, from this I can find out the poles okay and then write down the omega n and the damping ratio zeta right I am going to follow a slightly simpler technique okay which all of you know it so I am going to just write this in this terms you know which you all of you know I am just going to write the final result because I want to quickly get into the interpretation part so that I can write this down as a squared by omega n squared plus 2 zeta s by omega n plus 1 okay in this form write that last term in this form right from which okay when I rewrite that anyway that is equal to 0 so when I rewrite it in this form I will get omega n to be 2l divided by u root of c alpha f c alpha r into 1 plus k u squared divided by i z into m and zeta to be 1 by 2l into i z plus m a squared c alpha f i z plus m p squared c alpha r divided by root of i z into m c alpha f c alpha r into 1 plus k u squared done now we will get into why we did this. n will be 1 by I won't be a s squared you just see 2 zeta note that it is 2 zeta omega n so 1 by omega n squared l squared so l by okay I think right okay now why did we do this or what is the I hope I am just check quite simple just check whether it is okay now here we pick up a physical interpretation what is the physical interpretation of these things there are four quantities of interest one of the quantities I will write the expression later there are four quantities of interest in this one is what is let me call that as a1 all of this we have we have already derived okay 1 is what is called as the steady state response response of your velocity remember what this quantity was v divided by l into 1 plus k u squared which we just now wrote the second quantity of interest is omega n okay which is actually the natural frequency of the system the third quantity of interest is damping though remember calls this as natural frequency of yaw and natural frequency of damping I would resist from calling this because they are just the poles okay of the system I would not like to give them a name let us say that omega n is the natural frequency of the system and then let me introduce a fourth quantity which I would call as phi which is the phase lag 1 hertz of lateral acceleration okay now what is the physical meaning of the first one I know the expressions now okay so let us not worry about the expressions if I give you a system you will be able to find out so what is the physical meaning of steady state response of your velocity okay what do you think is the steady state response to your velocity what does it mean that is a ratio between your r to delta okay so higher it is what does it mean easily r would be developed or in other words you can interpret that to be the heading easiness I want to head how easy I am going to head okay so in other words how easily I am going to have the yaw for the car in order to get into the maneuver which I would like to have so this would give you what is called as heading easiness right okay now all of you know either natural frequency either you can express it in terms of radians per second or you can express it in terms of hertz okay all of you know something about the natural frequency of the system what is that you know about the natural frequency of the system you know or you understand the responsiveness of the system okay clear the larger the natural frequency remember that in your earlier classes for a simple spring dash pot system a second order system okay you were able to get what is the rise time what is the peak time TP what is the settling time TS okay overshoot all those things remember that you did those things for a second order system and that those were functions of 1 by omega n okay so omega n in other words gives me the responsiveness of the system okay heading for example now it is heading because that is what we use the word heading in order to understand the maneuverability so we get what is called as heading responsiveness of the system yeah what is what is the all the factors like settling time okay which is 4 divided by omega n to zeta okay and peak time all those factors are all functions of omega n okay larger the omega n smaller it is smaller is the same for example the peak time so smaller the peak time means I will reach I will reach what I want faster okay so it is ease with which I will go our response of the system is now better when omega n is increased right damping obviously all of us know it is it is responsiveness as well as there is a directional damping so I would not oscillate larger it is better that it would not get me an oscillating system right and the last thing is the phase lag what is the importance of 1 hertz is because most of the systems have you know the frequencies of excitation natural frequencies near 1 hertz so 1 hertz is just given but what is this phase lag what does that tell you what is what do you understand with the word phase lag or fun yes so in other words in other words intuitively I know that if the phase lag is small okay what what I give is immediately heard by the system or responded by the system or in other words it is easier for me to control the system okay so I would call this as controllability right so these are the 4 yes no this is for lateral acceleration phase lag of not so how do I find out find this out remember we had the transfer function yesterday okay in that transfer function we have to substitute s is equal to j omega and then you would get that at 1 hertz and that is what you would get now what murmur would date isn't this is this is not very difficult physics all of you know this what murmur would date was to plot this as a rhombus he plotted this as a rhombus so how did he do that in a very I would say interesting fashion okay. He had the natural frequency f n in one of the see there is no two axis look at this carefully the positive side he had f n he had phi increasing like this usually phi can be say minus e t minus 40 and so on okay that is how he plotted phi then he plotted the damping coefficient zeta in this axis right and of course the other one a 1 here why did he plotted like this very simple that if I now plot that for a car like this okay it simply means that understanding the physics of all these four it simply means that larger this rhombuses larger means f n is larger good for me phi is larger that means that look at how it is going minus 40 minus minus 80 minus 40 and so on okay it is usually the minus and then okay larger it is which means that your yes his response is very fast so that is a lag is small okay larger it is good for me damping and larger it is my first term which is the hidden easiness is better so he simply said that plot this like this when you plot it like this then the larger this rhombus the larger the rhombus better is the vehicle handling simple okay it is very easy to now understand or else it is not it is yeah you have to go into too much of details but here this is the first or the easiest way to understand vehicle handling look at the area of this rhombus and I am going to find out how good is my weight okay but but look at all the equations which are there may be some of them I have written here some of them you do not have here a 1 and 5 okay phase lag I have to write down that equation that is a tricky equation let me write down a 1 also so let me write down a 1 here a 1 which you already have u divided by l into 1 plus k u squared and phi is r tan a l c alpha r divided by l c alpha r u divided by 2 pi minus pi i z u that is a lengthy expression but I want you to check this slightly doubtful whether there is a 2 here or not fn divided by fn squared minus 1 just check this I think it should be right check whether that is correct now there is one small way small one issue what is the issue with all these expressions what is the issue the interesting issue here is that it not only depends upon the car of course it depends upon the tires extensively so the tires are going to play a great role in handling no issues okay in fact we have if you remember we have linearized the tire performance or tire characteristics and just put that initial slope okay not bad not the best way because we need to get this other tire models okay but we usually assume that we are we are not going to be away from the linear case we will come to how he determined this experimentally in a minute okay but before that let us look at these things carefully what are the other the peculiarities yes this and this has an effect more importantly velocities seem to have an effect okay the velocities are important so all these things are functions of velocity you are written down a yeah and velocities are important in other words a vehicle handling also depends upon the velocity so if you want to test a vehicle it is not enough if you test at 80 kilometers per hour or it is not enough if you draw this rhombus at 80 kilometers per hour so we also need to plot that for different velocities and what do you expect look at that carefully and tell me what would what would happen to a1 what would happen to other factors for a particular k most instances actually with increase in velocity v or u rather than our case as we increase the velocity rhombus now goes down so the words the handling characteristics depend upon the velocity vehicle which performs very well at 80 kilometers per hour okay will have will see a deterioration with velocities yes the typical value see because I have this formula okay we will do a problem in the during one of the assignments I have all this formula here okay I know the values of k just substitute it vary it with the velocity in fact if in other words if I want to now plot omega n versus I need to do say what I need to do is to omega n versus u if I now plot it the graph would look something like this clear so if I now plot each of these quantities with velocity for example omega n if I plot it would be like this so now I get those I have already the formulas there I get those and then plot it here right very simple nothing very difficult so now the other one is the understeer and the oversteered vehicle how is the understeered and the oversteered vehicle going to be affected now what would happen to an understeer I would as the vehicle becomes more and more understeered how will that go look at that formulas and tell me as the big vehicle becomes understeer what would increase okay omega n would increase a1 will decrease so it will move like this and obviously the oversteered vehicles move in the opposite direction now how did number determine for cars in other words what is the experiment he did or he recommended this group recommended an experiment called the pulse steer a pulse steer test in a pulse steer test in a matter of 0.4 seconds a steering is given an input okay of about 60 degrees and back so that is the pulse steer you give a pulse actually to the system okay so you give a pulse to the system it is like it is like giving an impulse okay so the pulse excites number of frequencies that is the reason why we went to the pulse of the system right here there are we can look at why pulse and so on from from f of t point of view but I am not going to do that this class but go back and read and maybe when you have time the later part of the course when we talk about ride and we talk about f of t we will look at the beauty of that pulse okay so he gave a pulse to system and said that at least about in 0.4 seconds okay this was done this was done so that the lateral acceleration or in the order of about 0.25 to 0.3 so that the assumption that we made regarding the tires that they are linear is valid clear in fact we have done a lot of work on this and we we found that this with an expert driver this rhombus okay which we call as memuro rhombus and the subjective evaluation for driver matched very closely so in other words we found this to be a excellent objective evaluation tool alright so now you know you see the importance of tires and we will see the importance of other things okay has to how they are going to be affected right any questions yeah depending upon the value of k this would vary and so I know the k values that you have come again what I suggest is that okay I understand that how much it would vary we will do a problem okay that is the easiest way to understand it okay how for example you are asking how big is the k okay would it be would it would it have an effect okay if k is larger small right we are talking about a typical car what is the k value which is an understeer coefficient car okay so what we will do is maybe we will do that in the next class okay I will work out and I will I will put some values to this so that you will understand how exactly that varies you know that is your doubt how much it varies we will put typical values in the next class and we will okay work out a problem that would give you an idea as to how it varies right in fact the experimental setup at experiment okay we will we will do that maybe in the next class this experiment you will do it in your vehicle dynamics lab class okay you will do this in the lab class so that is the reason why we there is a lot of experiments that need to be explained you know this course is too short we cannot do everything here so this course is more on theory the experimental vehicle dynamics you will do in vehicle dynamics lab right anyway we will pick up we will give you some dope on this in the next class okay I am going to give you an exercise okay do this you remember that we defined beta okay V by U remember that find out an expression for the transfer function for beta okay that will be an exercise and we will follow it up in the next class stop here and follow it.