Understanding Time Response for Control Systems

Ok. So, what we will look at today is something called the Time Response Specifications. So, last time what we had done was given a first order or a second order. system we just saw how it responds to different kinds of signals and in the first order system we saw there was something called a steady state error we will formalize that in a better way today. Second order systems we saw how the response system varies with changing in the damping term where we saw for under damped system your response is just continuous oscillation still you have critically damped systems to over damped systems and so on. So, what we will see now is given some design specifications what is the impact of the time response. So, loosely speaking if I say that I would want to go from. say you know some signal say y of t I want it to take some value of say y equal to 1 and somewhere here. So, ideally what if I would want to design a system I would want this guy to go instantaneously to this point at just say t equal to 0 plus that I switch on my air conditioner and immediately my temperature goes to the desired 24 degrees or 21 degrees ok. Sometimes my might be difficult to do it instantaneously, but it could be it could be possible to do it a little you know something like this right or something like this right ok. So, not exactly like this, but we could you know go you know in this time or this time, but what I would expect is if I you know want to run from this point till this point and do very fast then I just cannot stop here right I have to go a little further right before then I again come back and I run a little further. faster then I go here, here, here and so on right. However, if I would not want this thing to happen I would have to run a little slowly right so that I just go here and stop right. So, these are the kind of behavior we will analyze is to do with how fast can my system respond or how fast can it reach the given the desired value, but if it runs very fast are there any overshoots? And if there is any overshoot is there a limit on the overshoot or is a can I really say how say what is the value of y at this point given this guy to be 1 right or if you know when does it really reach the desired value right and so on. That is what we will try to you know quantify in this lecture. So and we will restrict ourselves to the step response of the system and the analysis becomes much easier and it is also you know quite intuitive too. So, in general all these specifications of how fast I go, how fast I settle down are a part of the design specifications. So, if someone gives me a problem would say design me something which reaches the steady state value in so and so time that it should not overshoot beyond a certain limits or this could be just 0.1 and I would say I would not want to overshoot 1.1 right for example. And these indices answer the following question. So, the based on this arguments what are we interested in. is how fast the system moves to a given input, is the response oscillatory and how oscillatory is the response which is again we know that it is depending on the damping that under damped systems will have an oscillatory response and how fast does it reach the final value right, ok. So, here we will restrict ourselves to under damped systems right and we will focus on the step response or the unit response. its step response. So, to quantify this we will define for ourself few things right. First is the delay time, the time required for the response to reach 50 percent of the final value at the first instance right. And I could calculate this based on the formula which I had derived earlier ok. I will come to this we just first define what these are and then go to see how these numbers are achieved. So, the delay time is a time required for the response to reach 50 percent of the final value at the first instance. You could have that in an oscillatory response that it could reach the 50 percent you know several times. Rise time is usually defined as the response to rise from 10 percent to 90 percent of the final value for over damped systems. And since we are here looking at under damped systems we will we will look at what time does it reach. the peak the desired value the first time or from 0 to 100 percent of final value at the first instance. For example, if I just were to plot it. So, this till this and I say this is my desired value these things right. So, here well this is what is called as the time right. So, where it reaches the value of 1 the first time you see that it reaches value here of 1 here, here and so on it keeps on you know oscillating and until it settles down to its final value. The delay time is this time right. So, this is a td where my response takes the value of 0.5. So, this is my time, this is my response y of t ok. Similarly, the peak time is the time required for the response to reach the peak value. So, somewhere here right. So, this is my peak time. Then, I will also quantify if I reach a peak if I am going from the desired value to some other value how much is this overshoot right. So, this is quantified or called as the peak overshoot and this. is just computed as the percentage of I am like. So, this is how much I deviate from the steady state value at the first instance. So, it is y t p which is what I overshoot here this one this is the value of y at t p minus the steady state value. then you just calculate that it in percentage of deviation from the steady state value. The settling time, so we had vaguely defined this last time when we were dealing with first order systems and we will recall that all over again. The settling time was defined as a time required for the response to reach and stay within a specified tolerance band. So, usually I would say well I am the system has reached some kind of a steady state value if it is within 2 percent of its final value sometimes even 5 percent depending on the depending on my tolerance levels. So, if I in this case where the steady state value is 1 I would say the settling time is when it reaches 2.98 right ok. So, again all these things. So, settling time we said we thought or we had derived the settling time concept earlier for first order system right. So, and these concepts the peak time and the peak overshoot are not defined for over damped and critically damped systems. That is because let us start with the response of. So, this is the under damped response, the critically damped response would be something like this right it will never cross this value of 1 or the desired value. This is the critically damped system. so this is the under damped system, this is under damped, this is critically damped. So, TP and MP are typically defined only for under damped systems and that is what we will focus most of our analysis on. So, this is what it looks like right. So, typical response of a first order system this we derived last time it looks like this is like has some kind of a damped oscillations right. So, I can have the delay time, the rise time, the peak time. and the settling time right. So, the first time when it reaches this guy. So, this could be within 2 percent could be 0.98 or even 1.02, but once with it is here it always remains within the 2 percent value right it will never go down. you know below 0.98 because the oscillations are continuously being damped. So, this is these are the values which we will which we will focus on T p, T r, M p and all this settling time. Now, the idea would be can I quantify this based on my system parameters right. So, start with the second order undamped system with transfer function which we had derived which we had said last time was. has a typical transfer function if I could call it g of s was omega n square over s square plus 2 zeta omega n s plus omega n square. And the response of this system and I will do this zeta is less than 1 and of course, greater than 0 right. So, the rise time. So, this system subject to a step input had a solution like this and we derived it last time that was ok with t not replaced by t r right. So, this t r is the reason we have t r is ok. So, the rise time is the time taken by the step response to go from 0 to 100 percent. So, this is is the 100 percent, this is the solution of the equation and I want to calculate the time t at which it reaches this value of 1 and therefore, I replace this t by t r in the solution to that equation. So, I do this the 1 and 1 goes away. So, I am left with sin omega d this is the damped national frequency if you remember was related to the undamped frequency in this way. sorry this was omega d ok. So, when this goes to 0 for the first time so which means omega d times t r plus theta equal to pi or t r is just this one pi minus theta over omega d and what was theta. So, theta is written here theta is cos inverse of zeta from the from the damping coefficient right, and then this relation between omega d and omega n the damp natural frequency with the with the natural frequency of the system. Similarly, I can calculate the peak time is the time taken by the step response to reach its peak value ok. Now, how do I find what is the time right. So, if I am here I am looking at this value. So, this is the function if I just say y is a function of t. So, it starts here, it goes here and again comes down right. So, this is the maximum value that the function can take and from calculus I know that the maximum. can be computed by taking the derivative of y right and then setting this to 0. This means I take dy by dt I set it to 0 and I compute the time at which this happens right. the time over here ok. So, I do all these computations. So, I take the derivative of this. So, this 1 disappears and I have is this one 0 is all these guys and then I solve this step by step to get this value that omega d times t p is this one and therefore, again I am just interested in the first instance right. So, this is one peak and then dy by dt would go to 0 here. by d t could go to 0 you know several points right here and so on, but I am interested only in the first time right this is the value which I would be interested in right. So, this T p would be simply pi over omega d ok. So, how do I express this overshoot as a percentage right. So, I calculate the peak value and then the steady state value is 1 the desired steady state value and here in this case it will just be m p would be be T p minus 1 right. So, how much I overshoot from here this number from here till here ok. So, I can just substitute y of T p here and I just y of T p would be just I know the value of T p that this is pi over omega t I plug it into the solution ok. This is the solution over here and and I just see how much it deviates from the steady state value of 1. So, y t p substitute for t p in the solution of y of t and this with all these computations would be something like this. So, m p is e power blah blah blah. theta and then this term gets cancelled because of cos of theta if you see here is defined this way right sorry theta is cos inverse of zeta theta is cos inverse of zeta or cos of theta is zeta and then I just use sin square theta plus cos square theta equal to 1 and this guy is just disappear. So, the peak overshoot in percentage is E. to the power you know this guy ok. Some properties of this one right. So, if I say well what are the limiting cases here right. So, we go from say zeta equal to 0 when there is no damping what we would expect is that m p would be the highest. So, if my steady state value is 1, then M p here would be substitute for zeta is 0 0 and so M p would be 1 which means my peak over overshoot would go from this is the value 0 to 1, this is the desired value it will go all the way till 2 and then come back and just keep oscillating like this. This is the standard response which we saw it is almost it is like a 100 percent overshoot. Now, if zeta keeps on increasing I see that my overshoot keeps on decreasing right. So, if I just go to plot. the peak overshoot versus zeta, the zeta going from 0 to 1 and see that this will go from 1 and all the way till 0 this is little important right. So, one of the design parameters here could be if I were to restrict my peak overshoot then. my problem would be to have to design my system with the appropriate zeta ok. So, similarly for settling time, the settling time is a time required for the response to reach and stay within say a 2 percent band of its final value ok. So, how do we compute this? So, let us say some pictures here that I am here and my response is again something like this right and this is my steady state value let us say this is 1 ok. We have to go a little further up here ok. So, if I look at this response here you can see it is actually bounded from below. So, by a curve like this and bounded from above by another curve like this ok. Now, what are these curves? These are simply these exponential curves here. The guy on the top would be 1 plus e power minus zeta omega n t ok. So, this r would just go over here, this will just be the standard t. over square root of 1 minus zeta square. So, this is the curve on the top and this is the curve on the bottom this is 1 minus e power minus zeta omega and t over square root of 1 minus zeta square and then. So, this value here would be 1 minus 1 over square root of 1 minus zeta square and the value here would be 1 plus 1 over square root of 1 minus zeta square ok. So, this is all good right. So, if I say well when does this curve inside the solid line reaches within 98 percent and stays there? Well the answer could be that whenever this curves you know this guy. whenever these reach the within the 98 percent and then stay there because these are enveloped by this curves right. This the guy on the solid line is just enveloped by these two curves. So, I can just now look at the response of this system right. So, I am just say let us take the bottom one here for example, 1 minus e power minus zeta omega and t over square root of 1 minus zeta square and this if you remember this looks like the response of a first order system right. Where we had the other day that the response was 1 minus e power minus t over square root of 1 minus zeta square and this is the response of a over tau right and if you just look at that response we had some things defined like this ok. This was my 1, this was my response y of t and this kind of look similar right. So, this is a time constant of t, this is a time constant 1 over zeta omega n and with t. So, this guy reaches t at t at the first time constant it reaches the value of 63 percent ok. And then at t equal to 2 times it reaches some value of 0.8 right and then at t equal to 3 t it reaches 95 percent and at t equal to 4 times the time constant it reaches 98.2 percent right somewhere here. This we had done this last time this 2 times t this is 3 times t this is 4 times t ok. Now if this this guy reaches settling time at 4 times the time constant, this guy would also be you know reaching in the same time right because this is just envelope by this by the curves there right. So, now my settling time can be defined. directly in terms of the time constant right. This is for the 2 percent tolerance band and settling time T s and then T s would be 3 times the settling time when I am at the 5 percent tolerance band ok. So, how do I write this? So, T s so if I again look at my curve the time constant there was say just take the bottom curve 1 minus e power minus zeta omega n t over 1 minus zeta square. So, this has a time constant of 1 over zeta omega n and therefore, my settling time would just be 4 over 4 times the time constant will be 4 over zeta omega n or the settling time this is for the 2 percent criterion and this would be 3 over zeta omega n for the 5 percent criterion. So, what does this mean that the settling time is inversely proportional to zeta and omega n. Again let us say the limiting case let us say the limiting case of zeta being 0 my settling time would be infinity. This is not surprising because if my settling time is infinity or when zeta is 0 I am just having like an oscillatory response right. My system will never go anywhere within the 2 percent or even the 5 percent or even the 50 percent right. So, settling time would be infinity. So, as zeta increases your settling time tends to reduce right. So, based on this formula here zeta increases and then the settling time reduces. So, I do not want to draw a graph for that at the moment, but we will just understand that the increase in zeta will cause a reduction in the settling time ok. Now, what is sufficient to increase zeta ok. So, my peak overshoot is also some function of zeta. ok. In such a way that if I have a smaller zeta, I have a smaller peak time or then even a smaller rise time right. So, if I want a faster response my zeta is smaller, but then my overshoot will be larger. So, usually in the design specifications my peak overshoot would be specified and my zeta would be calculated based on the peak overshoot. And therefore, if I were to look at settling time it just depends on the natural frequency of the system. to adjust settling time I can play around with this guy omega n right omega n also over here because zeta is already defined by the performance specifications of the peak overshoot ok. So, when this is fixed I may not be able to change it to here I can change it to here again because smaller zeta would maybe increase in a bigger overshoot which may not be very desirable ok. So, I can just play around with this omega n. And so, we will see you know what how we could smartly design these things so that I have a not a big overshoot, but also a smaller settling time and so on. So, ideally what we would like is to have faster response and a smaller settling time ok. So, what could be application of damped or this several kind of systems right. So, let us start with over damped systems. This is push button water tap shut off valve. So, just before we started this recording my student was telling me a nice analogy of this right. So, originally if you look at the trains the taps in the train we would have normal taps which people would just forget to turn off and then we could waste water. So, literally. on they had these things you know I do not know if I could draw a picture of this right. So, it is like this and then you will have a tap and then the water flows through here right and then you keep on pressing this and then once you release this the water will go will not come anymore right. So, what they do now is that you just press it. So, the system will you know give us water for some time and eventually the valve will close. Now that is like the example of an underdamped system like the emergency in technology is can be seen you know when you travel by train right. I do not know if the train still have this kind of smart things or not, but this is in more sophisticated thing you have the sensor you just put the hand and then it just actuates this and then the water just you know the valve just close off closes off gradually. Same thing you can see in this automatic door closers, they do not really be they cannot really have a fast response and then bang on to the wall or to the stopper right. So, these are automatic door closers are also typically over damped systems. Critically damped systems you would want them to have an elevator mechanisms or even gun mechanisms right. So, you want the elevator to have a faster response, but you would not want it to overshoot. So, we talked of this last time as well. And of course, under damped systems all string instruments would do that because you know if you plug a string say I have a guitar I plug the string and usually it has this nice wave kind of formation right. But if you plug the string from here it goes here and again comes back here it may not generate the sound which you want right. Similarly, all electrical or mechanical measuring instruments right. So, you take a voltmeter if it were to measure via a scale like this. So, if this is a 0 position it will typically go to the desired value and then come back and these are usually under damped systems right all this moving coil or moving iron measuring instruments ok. So, once. we do these things we are also interested does my output follow the reference all the time. So, we saw last time when we were doing the response for a first order system. system, in some cases we saw that there is a steady state error which means the reference is not tracked directly by the output, but it is with some error which was defined by the time constant right ok. So, let us try to formalize that a little more and then we will revisit the original examples which we had. So, the steady state error. So, a typical feedback loop looks like this you have the reference signal, the output, the feedback loop and the difference between the reference. interference and the output is termed as the error signal and this error I would typically want to go to 0. Now, given this characteristics of the system the g and this is 1 it will also be h nothing would change in the analysis. So, given this g can I find what is the error right depending on the nature of the signal here ok. So, what is the steady state error? It is error between the actual output and the desired output as time goes to infinity. So, we know from final value theorem if I call this signal as E of t that E s s is limit s going to 0 s times E of s in the Laplacian domain ok. Now, this E is r minus y and y I know is this transfer function here is g of s over 1 plus g of s ok. So, this y given as r minus g times r over 1 over g and I get this one ok. So, this is my error signal. So, in this thing the steady state error the limit s going to 0 s times e of s would be limit s going to 0 s times r plus 1 plus g ok. Now, this will make use of this little thing here and it will give us lots of information of the system subject to several inputs ok. So, that is all. So let us start with a step input right. So, this will be fixed all the time ok. So, for the step input R of s is 1 over s and what is the steady state error? E s s is limit s going to 0 s times R of s and this R of s is 1 over s right. So, this s and s disappear and I am left with this. So, this limit would be 1 over 1 plus k p right. So, this k p would be limit s going to 0 g of s and this limit will always exist. because we are we are looking at causal systems right. So, well does with this limit will always exist we do not really care about if this might blow up to infinity or not and this also we that this like the DC gain will always exist. So, where k p is computed as limit s tends to 0 g of s is called the position error constant right. So, I am just tracking a position a fixed position 1 over s. Now, similarly if I go to an input which is a ramp where I am tracking a ramp. So, this steady state error is I can I do all the computations one of the s disappears and I am just left with this this guy goes to 0 ok let us compute that. So, I have s limit s going to 0 s 1 over s square 1 plus s. So, this would be limit S going to 0 one of the S's would disappear and I have 1 over S plus S times g of S ok. Now, this would go to 0 this S and then what I am left with is limit S going to 0 S times g of S and I call this the k v. So, where k v is limit S going to 0 S times g of S and I call it the velocity error constant ok. So, if there is if I have a feedback loop I will just draw it here again. So, 1 reference plus minus y of s and so on right. So, if the input or the reference is a step then there will always be a error. So, the response so y of s. So, if I just take an under damped whatever kind of system and if I just look at the steady state curve. So, this is my desired value my actual value we might just be somewhere here right. So, this is the steady state error ok. So, we might think of so, the if this is the position error we might sometimes think that that the velocity error is actually that error in the velocity, but that is not true. It is just a velocity this is a error in it is not the error in the velocity, but error in position due to a ramp input or due to velocity input. For example, I am you know may be chasing a car here right. This is say a car C 1 and then this is my reference car and then I am just I just want to meet it right and at C 2 this is my car right. So, as time goes to infinity. So, we are always at say a distance of say 1 meter from this right. So, this is just error in the position we are not having error in the velocity even though we are just we could just be moving at constant velocities right. So, this could be some velocity v 1 this would also be the same velocity v 1, but I will just have a error in the position ok. So, thus it should not be confused with error in the velocity. Similarly, if I were to say well give me a temperature of 24 degrees, I just go to 23 the steady state error here here is 1 right. So, this I am just taking a position right I just want to go to 24, but however if I say keep on increasing my temperature profile with time in this way ok. So, say this is it increases let us say 1 times t ok. So, this is my t and this is my temperature the reference temperature right with time it just keeps on increasing. So, if I say that the velocity is you know there is a velocity error I start from here and I will just be here. So, if this has to increase constant. at t well I will always be behind by say 1 degree ok. So, at say at some t equal to 50 I might just be at 49 and t equal to 100 I might just be at 99 ok. So, it is just the difference in the position not really the rate of the rate of my car here or the rate of change of temperature here. Similarly, if I have a parabolic input for which the Laplace transform looks like 1 over S 3, I can do all the math and I say that the steady state error is 1 over k a which k a defined in this way. And, this guy is called the acceleration error constant and I just you know this has an extra s over here right ok. So, the error constants these three things and these are the standard test signals which we also use in the earlier lecture to study response of systems. So, the error constants k p, k v and k v they describe the ability of a system to reduce or eliminate steady state errors. For example, if I will come to that right. So, and then these values depend mostly on the type of the system where this type is important and as a type becomes higher more the steady state errors are eliminated ok. We will just define this type in the next slide right ok. So, I will come back to this slide little later, but just the type of a system is defined here by the number of poles at the origin. So, if I have a a system in the pole 0 configuration G of s could be typically some gain with a set of 0s with a set of poles and additionally some set of poles at the origin. So, if there are n poles I will call it a type n system, if n is 0 I will call it a type 0 system and so on right. And we will see how does this type actually now influence the way I compute the steady state error ok. So, that is the type of the system. So, the steady state error is a measure of how accurate my system is right, it does it really track my reference accurately or is there. So, the accuracy should so, in an ideal scenario I would always want it to track it to its desired value right. However, there could be situations where it may not be possible to do so. In that case I would like the error to be as small. as possible and that is a very important performance parameter right. I do not want a desired temperature of 24 and end up at 12 or 31 for example, right. So, this steady state error now depends on two factors right. It is the nature of the signal which I am tracking and also what is the nature of the physical system or the type of the system right. And of course, all these things we will only calculate for closed loop systems of course. I will calculate for closed loop systems only based on the open loop information I do not really need to see what is happening within the loop and so on ok. So, we will see why how we will do that. So, start for a type 0 system this g of s is again k set of 0s and there are no poles at origin right. So, this n in the previous slide this guy would be 0. So, I will just have these poles ok. So, how does it respond to a step signal? Well to for a step signal I have this guy 1 over 1 plus k p right. So, that there will always be some constant because some constant error be some constant error. at the steady state right. It can track position signal for example, this is my position signal this guy can track well possibly I would see the response, but could be something like this. and maybe it will go and then there will be some error right. So, there will be some steady state error here if I am just tracking this step. Now this system can never track a velocity signal why because the steady state error limit error is computed in this way limit s going to 0 1 over s times g s this is infinity. So, at steady state it will never be able to track a signal which is like this and the error will keep on increasing right. So, this guy as we keep on moving further the error will keep on increasing. Similarly, it will also never be able to track a parabolic input. So, the acceleration error will also be infinity. So, to summarize a type 0 system has a constant position error and infinite velocity. velocity and acceleration errors all at steady state. So, all these are related to this steady state behavior at the moment I am not worried about what is the peak overshoot, what is the rise time and so on. I am only interested in what or how my system behaves at the steady state ok. Type I system when I just do all the computations again I have one limit s going to 0. 1 plus g s and since this has a pole at 0 this guy will go to infinity and the k p. So, let us I can go back to see how we had defined k p. k p was defined at limit s going to 0 g times s. So, this guy since s has a pole at the origin this will go to infinity and the error will go to 0 ok. Similarly, the velocity error would be limit s going to 0 1 over s times g of s. So, this s. and this s would cancel out and I will have a constant here and the acceleration error will go to infinity. So, if I take a type I system and I ask it to follow a step input function. then it will at steady state there will be no error it will actually you know follow this input very nicely ok. However, if I take the same system I follow I ask it to follow a ramp it might have some steady state error right it might just go the response could be something like this and eventually go to a constant error here in the straight line ok. Similarly, acceleration it will never be able to track a signal which is it is parabolic. So, it will have a 0 position error. So, a constant velocity error and infinite acceleration error at steady state ok. Now, type 2 systems right. So, nothing will change here the position error will still be 0 because I have this s in the in the denominator I put it to 0 g of s goes to infinite. infinity and 1 over infinity goes to 0. Similarly, the velocity error. So, I do s times g s, 1 s still remains in my denominator. So, this will still be 1 over infinity and the velocity error goes to 0. However, my acceleration error constant. now has some number here it was infinity in the previous two cases. So, I could summarize by saying that if I take a type II system I ask it to track a position or a step it will track it perfectly right there will be no. steady state error. If I ask you to track a ramp it will also do it perfectly. If I ask you to track a parabolic signal it will do it, but there will actually be a small error determined by this acceleration error constant ok. Now we see that you know if I say give a little you know problem saying take a type 0 system and make the position error to be 0. What would I do? I will just say this looks remember all this theory now and I say just put an integrator right or add a pole at the origin. I can always do this is a system which has more poles than 0s I can possibly always realize this and I can just put it here right. So, this is can this be done, can this not be done or can this be done with some being a written. little careful we will see all those things, but these are the design specifications which would be given to us and then you know we can see that these errors can be quantified just based on the type of signals which we are tracking and the type of the system. So, having an integrator in my system or having a pole at the origin gives a little more hope to me in designing the system ok. So, in this entire module what we started with block diagram representations. signal flow graphs and then reducing the complexity or finding the transfer function by reducing the complexity of the block diagram or simply using the signal flow graph formula to find the transfer function. Then we also saw the time response of first and second order systems with standard inputs and we also justified why we were using this standard test inputs because in real time we can see all these signals the real time signals could be a combination of all these test signals. And then we actually quantified our performance in terms of the steady state error in in terms of how much do I overshoot, what is the rise time and so on and we had explicit formulas for those right. And in the next lecture we will and so far what we assume that the system is stable right and we had a very crude version of the definition of stability saying if it starts with 0 initial conditions it should come back to its original configuration. So, we will have notions of define these notions formally of stable and unstable system. systems and see how can we define these things directly by the transfer function looking at the poles and zeros just by looking at the poles and zeros can I say the system is stable or not or another notions where you know what is called as the bounded input bounded output stability. And then given a transfer function of fairly you know complex level which has lots of poles and zeros I will just teach you how we use the Routh Hurwitz criteria. for the stability of to determine the systems are stable or not ok. Thank you.

Ok. So, what we will look at today is something called the Time Response Specifications. So, last time what we had done was given a first order or a second order.

system we just saw how it responds to different kinds of signals and in the first order system we saw there was something called a steady state error we will formalize that in a better way today. Second order systems we saw how the response system varies with changing in the damping term where we saw for under damped system your response is just continuous oscillation still you have critically damped systems to over damped systems and so on. So, what we will see now is given some design specifications what is the impact of the time response.

So, loosely speaking if I say that I would want to go from. say you know some signal say y of t I want it to take some value of say y equal to 1 and somewhere here. So, ideally what if I would want to design a system I would want this guy to go instantaneously to this point at just say t equal to 0 plus that I switch on my air conditioner and immediately my temperature goes to the desired 24 degrees or 21 degrees ok. Sometimes my might be difficult to do it instantaneously, but it could be it could be possible to do it a little you know something like this right or something like this right ok.

So, not exactly like this, but we could you know go you know in this time or this time, but what I would expect is if I you know want to run from this point till this point and do very fast then I just cannot stop here right I have to go a little further right before then I again come back and I run a little further. faster then I go here, here, here and so on right. However, if I would not want this thing to happen I would have to run a little slowly right so that I just go here and stop right. So, these are the kind of behavior we will analyze is to do with how fast can my system respond or how fast can it reach the given the desired value, but if it runs very fast are there any overshoots?

And if there is any overshoot is there a limit on the overshoot or is a can I really say how say what is the value of y at this point given this guy to be 1 right or if you know when does it really reach the desired value right and so on. That is what we will try to you know quantify in this lecture. So and we will restrict ourselves to the step response of the system and the analysis becomes much easier and it is also you know quite intuitive too. So, in general all these specifications of how fast I go, how fast I settle down are a part of the design specifications. So, if someone gives me a problem would say design me something which reaches the steady state value in so and so time that it should not overshoot beyond a certain limits or this could be just 0.1 and I would say I would not want to overshoot 1.1 right for example.

And these indices answer the following question. So, the based on this arguments what are we interested in. is how fast the system moves to a given input, is the response oscillatory and how oscillatory is the response which is again we know that it is depending on the damping that under damped systems will have an oscillatory response and how fast does it reach the final value right, ok.

So, here we will restrict ourselves to under damped systems right and we will focus on the step response or the unit response. its step response. So, to quantify this we will define for ourself few things right.

First is the delay time, the time required for the response to reach 50 percent of the final value at the first instance right. And I could calculate this based on the formula which I had derived earlier ok. I will come to this we just first define what these are and then go to see how these numbers are achieved.

So, the delay time is a time required for the response to reach 50 percent of the final value at the first instance. You could have that in an oscillatory response that it could reach the 50 percent you know several times. Rise time is usually defined as the response to rise from 10 percent to 90 percent of the final value for over damped systems.

And since we are here looking at under damped systems we will we will look at what time does it reach. the peak the desired value the first time or from 0 to 100 percent of final value at the first instance. For example, if I just were to plot it.

So, this till this and I say this is my desired value these things right. So, here well this is what is called as the time right. So, where it reaches the value of 1 the first time you see that it reaches value here of 1 here, here and so on it keeps on you know oscillating and until it settles down to its final value.

The delay time is this time right. So, this is a td where my response takes the value of 0.5. So, this is my time, this is my response y of t ok.

Similarly, the peak time is the time required for the response to reach the peak value. So, somewhere here right. So, this is my peak time. Then, I will also quantify if I reach a peak if I am going from the desired value to some other value how much is this overshoot right.

So, this is quantified or called as the peak overshoot and this. is just computed as the percentage of I am like. So, this is how much I deviate from the steady state value at the first instance.

So, it is y t p which is what I overshoot here this one this is the value of y at t p minus the steady state value. then you just calculate that it in percentage of deviation from the steady state value. The settling time, so we had vaguely defined this last time when we were dealing with first order systems and we will recall that all over again. The settling time was defined as a time required for the response to reach and stay within a specified tolerance band.

So, usually I would say well I am the system has reached some kind of a steady state value if it is within 2 percent of its final value sometimes even 5 percent depending on the depending on my tolerance levels. So, if I in this case where the steady state value is 1 I would say the settling time is when it reaches 2.98 right ok. So, again all these things.

So, settling time we said we thought or we had derived the settling time concept earlier for first order system right. So, and these concepts the peak time and the peak overshoot are not defined for over damped and critically damped systems. That is because let us start with the response of.

So, this is the under damped response, the critically damped response would be something like this right it will never cross this value of 1 or the desired value. This is the critically damped system. so this is the under damped system, this is under damped, this is critically damped.

So, TP and MP are typically defined only for under damped systems and that is what we will focus most of our analysis on. So, this is what it looks like right. So, typical response of a first order system this we derived last time it looks like this is like has some kind of a damped oscillations right. So, I can have the delay time, the rise time, the peak time.

and the settling time right. So, the first time when it reaches this guy. So, this could be within 2 percent could be 0.98 or even 1.02, but once with it is here it always remains within the 2 percent value right it will never go down.

you know below 0.98 because the oscillations are continuously being damped. So, this is these are the values which we will which we will focus on T p, T r, M p and all this settling time. Now, the idea would be can I quantify this based on my system parameters right.

So, start with the second order undamped system with transfer function which we had derived which we had said last time was. has a typical transfer function if I could call it g of s was omega n square over s square plus 2 zeta omega n s plus omega n square. And the response of this system and I will do this zeta is less than 1 and of course, greater than 0 right.

So, the rise time. So, this system subject to a step input had a solution like this and we derived it last time that was ok with t not replaced by t r right. So, this t r is the reason we have t r is ok. So, the rise time is the time taken by the step response to go from 0 to 100 percent. So, this is is the 100 percent, this is the solution of the equation and I want to calculate the time t at which it reaches this value of 1 and therefore, I replace this t by t r in the solution to that equation.

So, I do this the 1 and 1 goes away. So, I am left with sin omega d this is the damped national frequency if you remember was related to the undamped frequency in this way. sorry this was omega d ok.

So, when this goes to 0 for the first time so which means omega d times t r plus theta equal to pi or t r is just this one pi minus theta over omega d and what was theta. So, theta is written here theta is cos inverse of zeta from the from the damping coefficient right, and then this relation between omega d and omega n the damp natural frequency with the with the natural frequency of the system. Similarly, I can calculate the peak time is the time taken by the step response to reach its peak value ok.

Now, how do I find what is the time right. So, if I am here I am looking at this value. So, this is the function if I just say y is a function of t. So, it starts here, it goes here and again comes down right.

So, this is the maximum value that the function can take and from calculus I know that the maximum. can be computed by taking the derivative of y right and then setting this to 0. This means I take dy by dt I set it to 0 and I compute the time at which this happens right. the time over here ok.

So, I do all these computations. So, I take the derivative of this. So, this 1 disappears and I have is this one 0 is all these guys and then I solve this step by step to get this value that omega d times t p is this one and therefore, again I am just interested in the first instance right. So, this is one peak and then dy by dt would go to 0 here.

by d t could go to 0 you know several points right here and so on, but I am interested only in the first time right this is the value which I would be interested in right. So, this T p would be simply pi over omega d ok. So, how do I express this overshoot as a percentage right. So, I calculate the peak value and then the steady state value is 1 the desired steady state value and here in this case it will just be m p would be be T p minus 1 right. So, how much I overshoot from here this number from here till here ok.

So, I can just substitute y of T p here and I just y of T p would be just I know the value of T p that this is pi over omega t I plug it into the solution ok. This is the solution over here and and I just see how much it deviates from the steady state value of 1. So, y t p substitute for t p in the solution of y of t and this with all these computations would be something like this. So, m p is e power blah blah blah.

theta and then this term gets cancelled because of cos of theta if you see here is defined this way right sorry theta is cos inverse of zeta theta is cos inverse of zeta or cos of theta is zeta and then I just use sin square theta plus cos square theta equal to 1 and this guy is just disappear. So, the peak overshoot in percentage is E. to the power you know this guy ok.

Some properties of this one right. So, if I say well what are the limiting cases here right. So, we go from say zeta equal to 0 when there is no damping what we would expect is that m p would be the highest.

So, if my steady state value is 1, then M p here would be substitute for zeta is 0 0 and so M p would be 1 which means my peak over overshoot would go from this is the value 0 to 1, this is the desired value it will go all the way till 2 and then come back and just keep oscillating like this. This is the standard response which we saw it is almost it is like a 100 percent overshoot. Now, if zeta keeps on increasing I see that my overshoot keeps on decreasing right.

So, if I just go to plot. the peak overshoot versus zeta, the zeta going from 0 to 1 and see that this will go from 1 and all the way till 0 this is little important right. So, one of the design parameters here could be if I were to restrict my peak overshoot then.

my problem would be to have to design my system with the appropriate zeta ok. So, similarly for settling time, the settling time is a time required for the response to reach and stay within say a 2 percent band of its final value ok. So, how do we compute this?

So, let us say some pictures here that I am here and my response is again something like this right and this is my steady state value let us say this is 1 ok. We have to go a little further up here ok. So, if I look at this response here you can see it is actually bounded from below. So, by a curve like this and bounded from above by another curve like this ok.

Now, what are these curves? These are simply these exponential curves here. The guy on the top would be 1 plus e power minus zeta omega n t ok. So, this r would just go over here, this will just be the standard t.

over square root of 1 minus zeta square. So, this is the curve on the top and this is the curve on the bottom this is 1 minus e power minus zeta omega and t over square root of 1 minus zeta square and then. So, this value here would be 1 minus 1 over square root of 1 minus zeta square and the value here would be 1 plus 1 over square root of 1 minus zeta square ok. So, this is all good right.

So, if I say well when does this curve inside the solid line reaches within 98 percent and stays there? Well the answer could be that whenever this curves you know this guy. whenever these reach the within the 98 percent and then stay there because these are enveloped by this curves right.

This the guy on the solid line is just enveloped by these two curves. So, I can just now look at the response of this system right. So, I am just say let us take the bottom one here for example, 1 minus e power minus zeta omega and t over square root of 1 minus zeta square and this if you remember this looks like the response of a first order system right.

Where we had the other day that the response was 1 minus e power minus t over square root of 1 minus zeta square and this is the response of a over tau right and if you just look at that response we had some things defined like this ok. This was my 1, this was my response y of t and this kind of look similar right. So, this is a time constant of t, this is a time constant 1 over zeta omega n and with t. So, this guy reaches t at t at the first time constant it reaches the value of 63 percent ok. And then at t equal to 2 times it reaches some value of 0.8 right and then at t equal to 3 t it reaches 95 percent and at t equal to 4 times the time constant it reaches 98.2 percent right somewhere here.

This we had done this last time this 2 times t this is 3 times t this is 4 times t ok. Now if this this guy reaches settling time at 4 times the time constant, this guy would also be you know reaching in the same time right because this is just envelope by this by the curves there right. So, now my settling time can be defined.

directly in terms of the time constant right. This is for the 2 percent tolerance band and settling time T s and then T s would be 3 times the settling time when I am at the 5 percent tolerance band ok. So, how do I write this? So, T s so if I again look at my curve the time constant there was say just take the bottom curve 1 minus e power minus zeta omega n t over 1 minus zeta square. So, this has a time constant of 1 over zeta omega n and therefore, my settling time would just be 4 over 4 times the time constant will be 4 over zeta omega n or the settling time this is for the 2 percent criterion and this would be 3 over zeta omega n for the 5 percent criterion.

So, what does this mean that the settling time is inversely proportional to zeta and omega n. Again let us say the limiting case let us say the limiting case of zeta being 0 my settling time would be infinity. This is not surprising because if my settling time is infinity or when zeta is 0 I am just having like an oscillatory response right.

My system will never go anywhere within the 2 percent or even the 5 percent or even the 50 percent right. So, settling time would be infinity. So, as zeta increases your settling time tends to reduce right.

So, based on this formula here zeta increases and then the settling time reduces. So, I do not want to draw a graph for that at the moment, but we will just understand that the increase in zeta will cause a reduction in the settling time ok. Now, what is sufficient to increase zeta ok.

So, my peak overshoot is also some function of zeta. ok. In such a way that if I have a smaller zeta, I have a smaller peak time or then even a smaller rise time right.

So, if I want a faster response my zeta is smaller, but then my overshoot will be larger. So, usually in the design specifications my peak overshoot would be specified and my zeta would be calculated based on the peak overshoot. And therefore, if I were to look at settling time it just depends on the natural frequency of the system. to adjust settling time I can play around with this guy omega n right omega n also over here because zeta is already defined by the performance specifications of the peak overshoot ok. So, when this is fixed I may not be able to change it to here I can change it to here again because smaller zeta would maybe increase in a bigger overshoot which may not be very desirable ok.

So, I can just play around with this omega n. And so, we will see you know what how we could smartly design these things so that I have a not a big overshoot, but also a smaller settling time and so on. So, ideally what we would like is to have faster response and a smaller settling time ok. So, what could be application of damped or this several kind of systems right. So, let us start with over damped systems.

This is push button water tap shut off valve. So, just before we started this recording my student was telling me a nice analogy of this right. So, originally if you look at the trains the taps in the train we would have normal taps which people would just forget to turn off and then we could waste water. So, literally. on they had these things you know I do not know if I could draw a picture of this right.

So, it is like this and then you will have a tap and then the water flows through here right and then you keep on pressing this and then once you release this the water will go will not come anymore right. So, what they do now is that you just press it. So, the system will you know give us water for some time and eventually the valve will close.

Now that is like the example of an underdamped system like the emergency in technology is can be seen you know when you travel by train right. I do not know if the train still have this kind of smart things or not, but this is in more sophisticated thing you have the sensor you just put the hand and then it just actuates this and then the water just you know the valve just close off closes off gradually. Same thing you can see in this automatic door closers, they do not really be they cannot really have a fast response and then bang on to the wall or to the stopper right. So, these are automatic door closers are also typically over damped systems.

Critically damped systems you would want them to have an elevator mechanisms or even gun mechanisms right. So, you want the elevator to have a faster response, but you would not want it to overshoot. So, we talked of this last time as well.

And of course, under damped systems all string instruments would do that because you know if you plug a string say I have a guitar I plug the string and usually it has this nice wave kind of formation right. But if you plug the string from here it goes here and again comes back here it may not generate the sound which you want right. Similarly, all electrical or mechanical measuring instruments right.

So, you take a voltmeter if it were to measure via a scale like this. So, if this is a 0 position it will typically go to the desired value and then come back and these are usually under damped systems right all this moving coil or moving iron measuring instruments ok. So, once.

we do these things we are also interested does my output follow the reference all the time. So, we saw last time when we were doing the response for a first order system. system, in some cases we saw that there is a steady state error which means the reference is not tracked directly by the output, but it is with some error which was defined by the time constant right ok.

So, let us try to formalize that a little more and then we will revisit the original examples which we had. So, the steady state error. So, a typical feedback loop looks like this you have the reference signal, the output, the feedback loop and the difference between the reference. interference and the output is termed as the error signal and this error I would typically want to go to 0. Now, given this characteristics of the system the g and this is 1 it will also be h nothing would change in the analysis.

So, given this g can I find what is the error right depending on the nature of the signal here ok. So, what is the steady state error? It is error between the actual output and the desired output as time goes to infinity. So, we know from final value theorem if I call this signal as E of t that E s s is limit s going to 0 s times E of s in the Laplacian domain ok. Now, this E is r minus y and y I know is this transfer function here is g of s over 1 plus g of s ok.

So, this y given as r minus g times r over 1 over g and I get this one ok. So, this is my error signal. So, in this thing the steady state error the limit s going to 0 s times e of s would be limit s going to 0 s times r plus 1 plus g ok.

Now, this will make use of this little thing here and it will give us lots of information of the system subject to several inputs ok. So, that is all. So let us start with a step input right.

So, this will be fixed all the time ok. So, for the step input R of s is 1 over s and what is the steady state error? E s s is limit s going to 0 s times R of s and this R of s is 1 over s right.

So, this s and s disappear and I am left with this. So, this limit would be 1 over 1 plus k p right. So, this k p would be limit s going to 0 g of s and this limit will always exist.

because we are we are looking at causal systems right. So, well does with this limit will always exist we do not really care about if this might blow up to infinity or not and this also we that this like the DC gain will always exist. So, where k p is computed as limit s tends to 0 g of s is called the position error constant right.

So, I am just tracking a position a fixed position 1 over s. Now, similarly if I go to an input which is a ramp where I am tracking a ramp. So, this steady state error is I can I do all the computations one of the s disappears and I am just left with this this guy goes to 0 ok let us compute that. So, I have s limit s going to 0 s 1 over s square 1 plus s. So, this would be limit S going to 0 one of the S's would disappear and I have 1 over S plus S times g of S ok.

Now, this would go to 0 this S and then what I am left with is limit S going to 0 S times g of S and I call this the k v. So, where k v is limit S going to 0 S times g of S and I call it the velocity error constant ok. So, if there is if I have a feedback loop I will just draw it here again. So, 1 reference plus minus y of s and so on right.

So, if the input or the reference is a step then there will always be a error. So, the response so y of s. So, if I just take an under damped whatever kind of system and if I just look at the steady state curve. So, this is my desired value my actual value we might just be somewhere here right. So, this is the steady state error ok.

So, we might think of so, the if this is the position error we might sometimes think that that the velocity error is actually that error in the velocity, but that is not true. It is just a velocity this is a error in it is not the error in the velocity, but error in position due to a ramp input or due to velocity input. For example, I am you know may be chasing a car here right. This is say a car C 1 and then this is my reference car and then I am just I just want to meet it right and at C 2 this is my car right.

So, as time goes to infinity. So, we are always at say a distance of say 1 meter from this right. So, this is just error in the position we are not having error in the velocity even though we are just we could just be moving at constant velocities right.

So, this could be some velocity v 1 this would also be the same velocity v 1, but I will just have a error in the position ok. So, thus it should not be confused with error in the velocity. Similarly, if I were to say well give me a temperature of 24 degrees, I just go to 23 the steady state error here here is 1 right.

So, this I am just taking a position right I just want to go to 24, but however if I say keep on increasing my temperature profile with time in this way ok. So, say this is it increases let us say 1 times t ok. So, this is my t and this is my temperature the reference temperature right with time it just keeps on increasing. So, if I say that the velocity is you know there is a velocity error I start from here and I will just be here.

So, if this has to increase constant. at t well I will always be behind by say 1 degree ok. So, at say at some t equal to 50 I might just be at 49 and t equal to 100 I might just be at 99 ok. So, it is just the difference in the position not really the rate of the rate of my car here or the rate of change of temperature here. Similarly, if I have a parabolic input for which the Laplace transform looks like 1 over S 3, I can do all the math and I say that the steady state error is 1 over k a which k a defined in this way.

And, this guy is called the acceleration error constant and I just you know this has an extra s over here right ok. So, the error constants these three things and these are the standard test signals which we also use in the earlier lecture to study response of systems. So, the error constants k p, k v and k v they describe the ability of a system to reduce or eliminate steady state errors. For example, if I will come to that right. So, and then these values depend mostly on the type of the system where this type is important and as a type becomes higher more the steady state errors are eliminated ok.

We will just define this type in the next slide right ok. So, I will come back to this slide little later, but just the type of a system is defined here by the number of poles at the origin. So, if I have a a system in the pole 0 configuration G of s could be typically some gain with a set of 0s with a set of poles and additionally some set of poles at the origin.

So, if there are n poles I will call it a type n system, if n is 0 I will call it a type 0 system and so on right. And we will see how does this type actually now influence the way I compute the steady state error ok. So, that is the type of the system. So, the steady state error is a measure of how accurate my system is right, it does it really track my reference accurately or is there.

So, the accuracy should so, in an ideal scenario I would always want it to track it to its desired value right. However, there could be situations where it may not be possible to do so. In that case I would like the error to be as small. as possible and that is a very important performance parameter right. I do not want a desired temperature of 24 and end up at 12 or 31 for example, right.

So, this steady state error now depends on two factors right. It is the nature of the signal which I am tracking and also what is the nature of the physical system or the type of the system right. And of course, all these things we will only calculate for closed loop systems of course. I will calculate for closed loop systems only based on the open loop information I do not really need to see what is happening within the loop and so on ok.

So, we will see why how we will do that. So, start for a type 0 system this g of s is again k set of 0s and there are no poles at origin right. So, this n in the previous slide this guy would be 0. So, I will just have these poles ok.

So, how does it respond to a step signal? Well to for a step signal I have this guy 1 over 1 plus k p right. So, that there will always be some constant because some constant error be some constant error. at the steady state right.

It can track position signal for example, this is my position signal this guy can track well possibly I would see the response, but could be something like this. and maybe it will go and then there will be some error right. So, there will be some steady state error here if I am just tracking this step. Now this system can never track a velocity signal why because the steady state error limit error is computed in this way limit s going to 0 1 over s times g s this is infinity.

So, at steady state it will never be able to track a signal which is like this and the error will keep on increasing right. So, this guy as we keep on moving further the error will keep on increasing. Similarly, it will also never be able to track a parabolic input.

So, the acceleration error will also be infinity. So, to summarize a type 0 system has a constant position error and infinite velocity. velocity and acceleration errors all at steady state. So, all these are related to this steady state behavior at the moment I am not worried about what is the peak overshoot, what is the rise time and so on. I am only interested in what or how my system behaves at the steady state ok.

Type I system when I just do all the computations again I have one limit s going to 0. 1 plus g s and since this has a pole at 0 this guy will go to infinity and the k p. So, let us I can go back to see how we had defined k p. k p was defined at limit s going to 0 g times s. So, this guy since s has a pole at the origin this will go to infinity and the error will go to 0 ok.

Similarly, the velocity error would be limit s going to 0 1 over s times g of s. So, this s. and this s would cancel out and I will have a constant here and the acceleration error will go to infinity.

So, if I take a type I system and I ask it to follow a step input function. then it will at steady state there will be no error it will actually you know follow this input very nicely ok. However, if I take the same system I follow I ask it to follow a ramp it might have some steady state error right it might just go the response could be something like this and eventually go to a constant error here in the straight line ok.

Similarly, acceleration it will never be able to track a signal which is it is parabolic. So, it will have a 0 position error. So, a constant velocity error and infinite acceleration error at steady state ok. Now, type 2 systems right. So, nothing will change here the position error will still be 0 because I have this s in the in the denominator I put it to 0 g of s goes to infinite.

infinity and 1 over infinity goes to 0. Similarly, the velocity error. So, I do s times g s, 1 s still remains in my denominator. So, this will still be 1 over infinity and the velocity error goes to 0. However, my acceleration error constant.

now has some number here it was infinity in the previous two cases. So, I could summarize by saying that if I take a type II system I ask it to track a position or a step it will track it perfectly right there will be no. steady state error. If I ask you to track a ramp it will also do it perfectly. If I ask you to track a parabolic signal it will do it, but there will actually be a small error determined by this acceleration error constant ok. Now we see that you know if I say give a little you know problem saying take a type 0 system and make the position error to be 0. What would I do?

I will just say this looks remember all this theory now and I say just put an integrator right or add a pole at the origin. I can always do this is a system which has more poles than 0s I can possibly always realize this and I can just put it here right. So, this is can this be done, can this not be done or can this be done with some being a written.

little careful we will see all those things, but these are the design specifications which would be given to us and then you know we can see that these errors can be quantified just based on the type of signals which we are tracking and the type of the system. So, having an integrator in my system or having a pole at the origin gives a little more hope to me in designing the system ok. So, in this entire module what we started with block diagram representations. signal flow graphs and then reducing the complexity or finding the transfer function by reducing the complexity of the block diagram or simply using the signal flow graph formula to find the transfer function.

Then we also saw the time response of first and second order systems with standard inputs and we also justified why we were using this standard test inputs because in real time we can see all these signals the real time signals could be a combination of all these test signals. And then we actually quantified our performance in terms of the steady state error in in terms of how much do I overshoot, what is the rise time and so on and we had explicit formulas for those right. And in the next lecture we will and so far what we assume that the system is stable right and we had a very crude version of the definition of stability saying if it starts with 0 initial conditions it should come back to its original configuration.

So, we will have notions of define these notions formally of stable and unstable system. systems and see how can we define these things directly by the transfer function looking at the poles and zeros just by looking at the poles and zeros can I say the system is stable or not or another notions where you know what is called as the bounded input bounded output stability. And then given a transfer function of fairly you know complex level which has lots of poles and zeros I will just teach you how we use the Routh Hurwitz criteria.

for the stability of to determine the systems are stable or not ok. Thank you.

Transcript for:Understanding Time Response for Control Systems

Transcript for:
Understanding Time Response for Control Systems