 Welcome everyone to this lecture. Today we will start with a new topic called describing functions. This technique has widely been used since perhaps 1930s, 1940s or a little later and it is a method for finding out periodic orbits that is where we will see. It is an approximation method. So, the periodic orbit values, the amplitude and the frequency, those values are expected to be approximate. So, more precisely we can say to obtain likelihood, likelihood meaning it is not a method that guarantees easily. There are sufficient conditions for guaranteeing periodic orbits and also sufficient conditions for ruling out existence of periodic orbits, but otherwise what we will study in this course is more about the likelihood of periodic orbits and amplitude frequency. This is something in continuation of what I told in the very start of this course that we need non-linear systems for sustained oscillations for robust sustained oscillations. Keeping that in mind, we will see an example of how non-linear systems can achieve this and the technique used for finding the amplitude and frequency goes through this particular method called describing function method. This describing function method is not to be confused with another word. There is something called descriptor which is also called a singular system. This one is a linear system. So, we have all seen state space LTI system. This is LTI system in which E is singular, singular matrix. Determinant of E is equal to 0. This is descriptor system. This descriptor is not the same as describing function. Descriptor system is not the same as describing function. When I started studying describing function many years ago, many places on the net had interchanged these words. So, please read very critically. Whatever you read on the web, one should go for established authors, textbooks or research papers by established authors rather than relying on the net. If one wants to use the net, of course one should because so many things there are all free. That time one should read everything very critically. So, please note that descriptor system is different from describing function. This is a linear time invariant system while this is an approximation method in non-linear systems. Approximation method for finding out periodic orbits, their frequency and amplitude. So, let us take a specific example. Let us begin with a specific example. So, consider a system G, a transfer function, a linear system which has in the feedback path the saturation non-linearity. So, this particular device, suppose this output is input to this saturation non-linearity is y and output is u, then the graph of this is up to some value. It is linear, outside some value it saturates. We will call this saturation non-linearity. For making this standard, we can make this equal to 1, this equal to 1, this is minus 1, this is minus 1. So, it is linear in the range minus 1 to 1, that time it has slope 1, outside that range it has saturated to that value plus 1 here, minus 1 here. So, this is an example of a memory less non-linearity. What is memory less about it? The output to this device, the output u depends only on the value of y now. It does not depend on the rate of change of y, does not depend on the second derivative of y. In other words, whether y is increasing or decreasing, it takes this value. Further, this dependence of u on y does not depend on time explicitly. It is memory less and time invariant. Many of the commonly encountered non-linealities are like this. Memory less and time invariant non-linearity. Another example is the dead zone. This is a dead zone. Because we have this non-linearity in the feedback path, we prefer calling input to this non-linearity as y and output is u. So, that output of the original system continues to be y. So, this is another example of a memory less time invariant non-linearity. This is called dead zone. It can also be understood as play in the system. This is more correctly called as dead zone non-linearity. This is saturation non-linearity. These are extremely widely encountered in practice. So, let us, if one wants to increase the slope, if one wants to change the slope of this, all one can do is just multiply this output u by a constant k, so that the slope for this range is equal to k. So, that can easily be accommodated. Let us see how it can be accommodated. So, let us take this example of a g of s in the feedback path. It has the standard saturation non-linearity. Standard is not a very standard term. I am calling it standard to say that it is exactly slope 1 in the range plus and minus 1 and then it is saturating plus or minus 1. Saturation, what comes as output goes through a linear block k. So, one can easily consider an amplification by k and then by such an amplification, one can get any other slope also. If you want a slope of k in this range, if you want a slope of, if you want slope of value k, one can easily make this change. By that, you can get this by just making this output go through a constant gain called k. So, let us look into whether there can exist periodic solutions. Let us take an example, g of s equal to 1 over s plus 1, s plus 2, s plus 3. Let us take this particular transfer function and see for what value of k one might expect periodic orbits. For doing this, we will use some linear systems theory. We will use a development of Nyquist criteria, Nyquist plot and Nyquist criteria for stability, which is all applicable only for linear systems from that we will find a value of k, which we will, beyond that value of k, we expect that there will be periodic orbits, there will not be instability. That is what we will calculate in detail now. So, as I said, we will first calculate for a linear time invariant system in which g of s equal to 1 over s plus 1, s plus 2, s plus 3. Find value of k for stability. This is the problem. So, what one can do is, one can plot the Nyquist plot of g. Nyquist plot of g with this negative unity feedback will give, and depending on the number of encirclements of the point minus 1, we can get lot of information about whether the closed loop is stable for unity gain and even if it is stable for unity gain, for what value of gain in the feedback path, it will become unstable. See, it is only the loop gain that matters. This negative sign is important. This k could have been either in the feedback path or in the feed forward path. So, let us consider k in the feedback path now. So, this Nyquist plot, if it goes like this, what is this value? It is 1 by 6. Why 1 by 6? Because this is where g of 0, where omega equal to 0 starts. This is omega equal to 0, then it becomes like this and this is where for omega equal to infinity and then from omega equal to minus infinity onwards, it comes like this. So, we are interested in calculating this point precisely where it intersects the negative real axis because for larger values of k, we expect that it will encircle this point minus 1. So, first let us find out whether it is whether the closed loop is stable. Of course, it is easy to infer that the open loop is stable because the poles are all nothing but minus 1, minus 2, minus 3. All the 3 poles are in the left half complex plane. That does not mean that the closed loop will also be stable. Closed loop is stable by that we mean the transfer function from r to y. For that purpose, we are interested in finding the number of encirclements of the point minus 1 by the Nyquist plot and the number of open loop unstable poles. So, we will quickly review the Nyquist criteria for stability also. So, we have the Nyquist criteria of stability as count number of anti-clockwise encirclements of point minus 1. So, we are speaking y minus 1 because we have negative unity feedback. Let us take h because that is more standard in various textbooks. So, consider this feedback. We are going to plot the Nyquist plot of G h actually. So, first we will review the Nyquist criteria for stability for this case and then we will replace h by k. So, plot the Nyquist plot of G h on the complex plane and then count the number of anti-clockwise encirclements of the point minus 1 and call this integer as n. If the encirclement is clockwise, then you count the number and just put minus sign to it. So, this integer n can be positive or negative. If it is positive, then it has indeed the Nyquist plot has encircled upon minus 1 anti-clockwise. If this n is negative, that would be because we have clockwise encirclements of the point minus 1 and y is minus 1 important because we have negative sign here. That is why minus 1 is important. Then we also have two more integers. P is number of unstable open loop poles, open loop poles of which transfer function of 1 plus G h actually. 1 plus G h is the important thing and you see if you add a number 1, the poles of this is same as of G h which is a loop gain. The poles do not change by just adding a fixed number. If you add a fixed number, of course, zeros might change that is indeed important, but the number of unstable open loop poles of 1 plus G h and G h are the same. That is because it is of G h, this open loop should be replaced by loop gain. If it is constant gain K in the feedback, then this is nothing but G in that case. So, it is number of unstable poles of 1 plus G h is how the Nyquist criteria, Nyquist plot, Nyquist criteria reads, but it is exactly loop gain because unstable open loop, the number of unstable poles do not change by adding a constant and when the feedback path is a constant either K or 1, then it is nothing but number of unstable poles of G and hence it is also called the number of unstable open loop poles. Open loop poles that is before you close the loop. If you close the loop with a dynamic transfer function called h, then of course you should be counting the number of unstable poles of G h which is same as 1 plus G h and why is 1 plus G h important? Because 1 plus G h comes in the denominator in the transfer function from R to Y. Consider this. So, Z is the eventual unknown. We are using Nyquist criteria for stability to obtain this number Z. This is number of unstable zeros of 1 plus G h. This is the most important index integer in the three things. It is to find Z that we are trying to use Nyquist criteria for stability. Z is used because of it is zeros, but zeros of which that is where many, many books. In fact, I have seen textbooks also that give this particular criteria wrongly. It is number of unstable zeros, but not of open loop transfer function G, not of loop gain G h, but of 1 plus G h. This it is number of unstable zeros of this which is actually unknown. It is unknown to whom now? It is unknown in this Nyquist criteria for stability. It is unknown that is what I will first clarify. If these are three integers, then Nyquist criteria for stability says n is equal to P minus Z. We are supposed to use the Nyquist criteria for stability this particular equation. This equation, there are three variables, only one equation. So, in three variables one equation, we expect two to be known values so that we will use the equation to find out the third variable which can be unknown. So, n comma P known and Z unknown. Unknown means what? Use equation to obtain. So, please note that this is extremely important. This is often told wrongly in various books. So, Nyquist criteria for stability correctly is like this. It is n is equal to P minus Z. Three integers get related by the Nyquist plot. What are these three integers? n is the number of anticlockwise encirclements of the point minus 1. Anticlockwise refers to that these encirclements should be counted positive. If it is anticlockwise, it should be counted negative if it is clockwise. P is the number of unstable poles of 1 plus G h and Z is the number of unstable zeros of 1 plus G h. So, why is Z important? Z is number of unstable zeros of 1 plus G h, that is what I have explained just now. But why should we be concerned about the number of unstable zeros of 1 plus G h? That is because transfer function from R to Y is nothing but G over 1 plus G h. So, the zeros of the denominator are precisely the poles of the transfer function from R to Y. So, please come back to this particular figure. In this figure, the transfer function from R to Y, the transfer function is nothing but G over 1 plus G h. One can obtain this by very straightforward block diagram reduction procedures. In this transfer function from R to Y, 1 plus G h comes in the denominator and if you do not want unstable poles in the transfer function from R to Y, then this denominator 1 plus G h should not have unstable zeros and it is to count unstable zeros of 1 plus G h that we are using the Nyquist criteria for stability. So, we will conclude that the closed loop is stable if Z is equal to 0. Please do not confuse Z as the number of zeros, number of unstable zeros of G nor of G h. Z is the number of unstable zeros of 1 plus G h, that is why Z, but it is the number of unstable closed loop poles. There, when you are speaking of the closed loop, that time Z indeed denotes the number of unstable closed loop poles. So, it is to obtain Z that we are doing this exercise. N and P are known. We will use G, knowledge of G and H to find out P. We will plot the Nyquist plot and obtain N. N refers to the number of anticlockwise encirclements of the point minus 1. Minus 1 point significance comes because of this negative sign in the unity feedback. Because of this negative sign, we have got 1 plus G h as the denominator. That is how we will find, using the Nyquist plot, we will find P and then we will use this equation, plug in N and P and obtain Z. If Z is equal to 0, we will conclude that the closed loop is stable. Closed loop is stable, no unstable closed loop poles, which is same as no unstable zeros of 1 plus G h, which is same as Z equal to 0. We want to conclude closed loop is stable or not. For that, that is same as saying that the closed loop has zero number of unstable poles, which is same as the number of unstable. This n o stands for number, n o stands for no unstable zeros of 1 plus G h. 1 plus G h comes in the denominator. Hence, we are interested in the zeros of 1 plus G h and this Z stands for zeros, but not of G h, but of 1 plus G h. So, closed loop is stable. In other words, if N is equal to P. So, this is the famous closed loop is stable if N is equal to P. If the open loop has unstable poles, open loop G is all of concern if h is a constant. Otherwise, you should be counting the number of unstable poles of the loop gain G h. So, if that P is not equal to 0, then you had better have that many anticlockwise encirclements of the point minus 1 in order to get the closed loop to be stable. That is how the Nyquist criteria for stability should be read as. If the open loop is already stable and if you have only constant gain feedback in the feedback path, then P is equal to 0, then indeed you want N equal to 0 if you want closed loop stability for that constant gain in the feedback path. So, coming back to our example, we had been working on the specific transfer function G of S equal to 1 over S plus 1, S plus 2, S plus 3. For this transfer function, I already drew, I already sketched that this is 1 by 6. It starts from here. Since we have only poles, the body plot is expected to only have its phase plot decreasing. So, which means that the Nyquist plot is going to go on coming closer to the origin because the magnitude is also falling and also the phase is decreasing from 0 eventually to minus 270 that is when it reaches 0 amplitude and for the other part is just reversal. Point minus 1 we expect is very far. Why do we expect it is very far? Because it has started at 1 by 6 here and the magnitude is only falling which means that it is coming closer and closer to the origin. So, it cannot have crossed minus 1, but nevertheless it is very important to know at which point it intersects the negative real axis that is what we will calculate quickly. So, we will say find omega value where G of j omega is real which means that imaginary part of G of j omega equal to 0. One might constantly ask why are we doing this for describing function method, but I should tell you that this same example we will use for describing function. We will use it for sector non-linearity calculations also for circle criteria. So, for all these purposes same example will serve a very key role and of course we have reviewed a very important topic in linear system theory namely the Nyquist criteria for stability. So, we expect that for gain larger than some value even for saturation non-linearity we expect periodic solutions. We will quickly see that it cannot be unstable hence it has to be periodic. So, we are going to now calculate those values of omega where the imaginary part is equal to 0 and for that value of omega when we substitute we will get the real part that will give us this real axis intersection. One of course we expected 1 by 6, the other values if any will be on the negative real axis this is what we will conclude very soon by equating imaginary part of G of j omega to be equal to 0. So, G of j omega is equal to 1 over j omega plus 1 j omega plus 2 j omega plus 3. So, which is nothing but equal to 1 over why do not we expand one these two brackets together, multiply it inside this gives us minus omega square j omega there is 2 j omega coming from here and this this gives 3 j omega plus 2 this remains as j omega plus 3 this is equal to 1 over please do this bracket opening very slowly one can easily make mistakes and that will affect all the conclusions later. So, mistakes here are kind of unpardonable because we have to keep in mind that the consequences because of small calculation mistakes are very bad later hence more care had better be given here no excuses for not giving care here. So, this gives minus j omega cube here then we have omega square term coming from minus 3 omega square here and again minus 3 omega square here. So, that that gives minus 6 omega square then we have the j omega the j omega how many of them is what we have to calculate. So, j omega comes 2 j omega come from here and 3 and 9 j omega come from here that makes it 2 plus 11 j omega and finally plus 6. So, if we have calculated correctly then this is equal to this. So, imaginary part of j of j omega equal to 0 means you see the numerator is real denominator is all that is making this whole number a complex number. So, if the imaginary part of the whole j of j omega should be equal to 0, imaginary part of the denominator should be equal to 0. So, imaginary part of this equal to 0 is if and only if minus j omega cube plus 11 j omega equal to 0 in other words omega equal to 0 or omega square equal to 11. These are the two situations that cause the imaginary part of the denominator equal to 0. Hence, these are the only two situations which can cause the entire j of j omega to be a real number. So, at omega equal to 0, j of j omega is equal to 1 over 1 over what? In this part we expect that the imaginary part do not have to be evaluated because they will all cancel off that that should indeed be checked once again, but please do this part of the calculation yourself. We are going to evaluate only the real part. So, we are going to get 6 minus 6 into 11 which is equal to 1 minus 1 over sorry I this is that omega equal to square root of 11. This is what we have evaluated at omega equal to 0 j of j omega equal to that is easier at omega equal to 0 we get this is equal to 1 by 6. So, these are the only two places omega is equal to plus minus square root of 11 and omega equal to 0 are the only three places where we have real axis intersection. One of these cases is for the positive real axis. The other two intersections happen to be at the same point that is expected because we expect that it is symmetric with respect to the real axis and hence if one part of the plot for omega positive if it intersects at some point for omega negative also it will intersect at the same point. So, this particular point we have now inferred is equal to minus 1 by 60. We knew that this will be closer to the origin than 1 by 6 because the Bode magnitude plot is only decreasing because of only poles. So, to have only poles means it is going to be a low pass filter. So, it starts at 1 by 6 and goes on continuously decreasing to have any poles to have only real poles I should correct myself. If one has only real poles and no pole at the origin that is a situation where the magnitude can only keep decreasing. So, the next plot from wherever it starts will keep coming closer and closer to the origin. So, these are the points where it has intersected the negative real axis. So, what does this minus 1 by 60? What is the significance of this? That is what we will see now. So, if we have, so now let us by a small manipulation put the k up there. As I said it is only the loop gain that matters. If we have k here instead of 1 what we already concluded is that for unity feedback which is nothing but k equal to 1, k is equal to 1 in either this configuration where k is in feedback path or k is here in the feed forward path. In both these cases notice that the loop gain is the same and it is only the loop gain that matters as far as stability is concerned. The transfer function from R to Y is of course affected whether k is in feed forward path or feedback path but the closed loop stability conclusions do not depend on where inside the loop these k and g are located. As long as they are all in the loop they all give the same closed loop stability conclusion. So, coming back to this figure for unity feedback for k equal to 1 we have already concluded that the Nyquist plot does not encircle the point minus 1. This is minus 1, this is 1 by 6, this we have said minus 1 by 60 even though I had already told that it has to be to the right of the point minus 1. We have in fact calculated the point precisely it is at minus 1 by 60. So, we know that for unity feedback equal to 1 n is equal to 0, p is also equal to 0 because g is stable open loop stable g is equal to 1 over s plus 1, s plus 2, s plus 3. So, all the poles are in the left half complex plane. So, the loop gain g and hence k g also both have p equal to 0. Hence, now we plug this all into the formula n is equal to p minus z this gives us z equal to 0. So, for unity feedback that is for k equal to 1 we have concluded that the closed loop is also stable in addition to the open loop. But now we go on multiplying this k for k equal to 2, 3, 4, 5. Notice that if we have done so much effort in plotting the Nyquist plot of g plotting the Nyquist plot of 5 times of g just means that we multiply this Nyquist plot by 5 and it blows up. To multiply by any positive number means just scaling and making this Nyquist plot larger to multiply the Nyquist plot by minus 1 does not mean we reverse this like this. But what it means is we rotate this by 180 degrees to multiply by minus 1 does not mean that we swap this like this. We mirror image this with respect to the image axis but to multiply by minus 1 in fact means to rotate this whole plot by 180 degrees either plus 180 or minus 180 that is immaterial. That indeed makes a difference though you see the whether it is rotating a point by in a clockwise anti-clockwise orientation that is going to get reversed if we mirror image this plot. But whether it is rotating clockwise anti-clockwise that does not change, orientation of the plot does not change by rotating this by rotating the entire plot by 180 degrees. So please note these are extremely different things. When you multiply by a minus 1 then keep note to rotate the plot by 180 degrees rather than to mirror image the plot about the image axis. So now what we are able to say is for k larger than 1 or for k positive for larger than 1 this plot only becomes larger and larger and it will not encircle the point minus 1 until you multiply by k equal to up to 60. For k less than 60 n is still equal to 0 and by whatever k you multiply of course the number of unstable poles of kg do not change by multiplying by a constant k. So p continues to be 0 and hence z equal to 0. But what happens when you put k larger than 60 for k larger than 60 if you multiply this Nyquist plot by k larger than 60 then this this Nyquist plot would have become large enough so that this point of intersection on the negative real axis comes to the left of the point minus 1. So let me take an example. Let me take say k is equal to 65. Let us take an example and for the ease of understanding let us explicitly plot it for this situation. Nyquist plot of g of s equal to 65 over s plus 1 s plus 2 s plus 3. Are we going to do all the calculations again? Of course not. We have only plotted for 1 over s plus 1 s plus 2 s plus 3. All we are going to do is multiply it by 65. So we have got that it starts from 65 by 6 and slowly comes inside. We do not know where the point minus 1 is but notice that this point is equal to now minus 65 by 60. Nyquist plot is useless without the orientations especially now that the minus 1 point is getting encircled. How did we conclude that the minus 1 point is getting encircled? This negative real axis intersection has been calculated as 65 times minus 1 by 60 which is nothing but minus 65 by 60 which is slightly less than minus 1 which concludes that the point minus 1 is inside here. So now we can count how many times it is encircling this point minus 1. Where is the point minus 1? It is here. So it has encircled clockwise. So n is equal to 2. It has encircled the point minus 1 twice but clockwise. That if it is clockwise then we are supposed to take this number 2 as negative minus 2. p continues to be 0 and hence now we are going to put n is equal to p minus z. So we have got z equal to 2. So one of the best things of Nyquist criteria is that everything fits well like magic. For no Nyquist plot will you ever get z negative? It is not possible to say that the number of unstable right half, the number of unstable close to poles is minus 2. We can have negative number for n. For p and z we have to have positive numbers only. So if you have p is equal to 0 and if you have instability then the encirclement of the point minus 1 has to be clockwise. All these things are so well rigid. They are also interlinked that one part ensures the other part so well. So I am trying to say that you will never encounter a situation where z has become negative. So what it means is there are two unstable poles for k larger than 60. This is the conclusion for any larger value of k also. So for k greater than 60, n equal to minus 2, p continues to be 0 of course hence z equal to 2. We have two unstable poles of the closed loop for any value of k greater than 60. This is how one can use Nyquist plot for also finding the precise range of k for which we have closed loop stability just like one can use Ruth Hurwitz criteria also. So now let us come back to our saturation nonlinearity and see what this says for the saturation nonlinearity. So we have a constant gain k here and k is equal to let us say 65. This is y. We have 2, we have k here. We are going to analyze two systems closely. Here also r equal to 0. So now here we know that for k is equal to 65 we have closed loop stability even when r equal to 0 to have instability means that if the initial conditions are not equal to 0 for very small perturbations we will have oscillations building into the system, oscillations that slowly grow until things become unstable. So this is how y looks as a function of time y of t. If it starts from nonzero initial conditions then we have instability, oscillations that slowly grow. In other words, they cannot be sustained. These oscillations are slowly going to grow and become very large. Why? Because two poles are in the open right half complex plane for k is equal to 65. But when these oscillations are very small, when y is very small then we expect that it is inside the range plus minus 1. So they will continue to grow inside this range. As long as the output is inside this range it will grow. But the output cannot, after all y is limited what comes out here cannot be greater than plus minus 1. And it is getting amplified by only 65. Do we really expect that oscillations will become unbounded in this example? Here because of this saturation in the feedback path the oscillations cannot become unbounded. Does that mean that they will all die down to 0? No, they cannot die down to, as soon as the oscillations are smaller than this range as long as y is less than plus minus 1 there will always be a tendency to grow because if y is 4, 4 minus 1 less than y of t less than y less than plus 1. As long as y is inside this range the input here, the y input to this particular saturation nonlinearity block looks at this block as a linear system. And hence what comes out will only want to grow as it goes like this. Why would it want to grow? Of course this is just gain 1. For y inside this range this is just gain 1. But why would it want to grow? Because in that case it is like this system where you have 1 in the feedback path. And for k equal to 65 we know that this is unstable. So it will want to grow because of the dynamics of k and g together and the feedback aspect of this configuration. It will want to grow. But once it grows beyond plus or minus 1 it is getting chopped, it is getting saturated. So we cannot have a situation like this here. So we have to have oscillations, we have concluded and oscillations are sustained. What is the meaning of sustained? Don't die to 0. Why they do not die to 0? Let us just recapitulate. It doesn't die to 0 because as soon as the range of y, the oscillations here come smaller than plus minus 1, this block behaves like a linear system with unity gain in the feedback path. That time it is like this system and we know that this is unstable. So the oscillations can't come smaller than plus minus 1. The amplitude will be greater than 1, cannot become unbounded. Why it cannot become unbounded? Because we know that there is a saturation feedback path. At this point the gain can never exceed the value of the signal here, can't exceed 1 in magnitude and this is at most 65, this is at most 65 magnification, r equal to 0 in both the cases, external input is 0. We are looking at oscillations of this autonomous system. So that is why we know that this system can't have unbounded trajectories either. It cannot become, oscillations can't become unbounded. In other words, it's sustained. This is an example of an oscillator where we have sustained oscillations but of course this is not a linear system. We need more techniques to find out the amplitude and frequency of the signal of the oscillations here. Of course at least the frequency is not hard to find, at least approximately but exact values are a very complicated task. We will use describing function method to find out the amplitude in particular. For this example, for k equals 65, 70, for different values, we would like to find out the values of the amplitude. This is an example where describing function method will help. So, what exactly is a describing function? We saw some motivation for describing function for finding out, some technique is required for finding out the gain and the frequency when we have a non-linear element in the loop, in the feedback loop. So, one should think of the describing function as a gain. It is a function of, it is a gain. We are going to denote it by eta, a gain that depends on the frequency and amplitude. So, for every block, linear or non-linear, we prefer calling the input as Y, output as U only because in our applications we will have this non-linear element in the feedback path. For example, it could be a saturation non-linearity which we have seen or it could be a dead zone or it could be a dead zone like this. So, corresponding to every block, we want to associate a notion of gain of this operator. And we are going to keep in mind that this gain can depend on the frequency omega, it can also depend on the amplitude. What are these amplitude and frequency? This is when the input Y is equal to the signal Y of t equal to A sin omega t. Now, we need a little more development before we reach the stage where we define the describing function as a gain. It is a complex gain. It has a real part and an imaginary part, but the motivation of this is completely linear time invariant systems. Why? Because for linear time invariant systems, the transfer function of the system is nothing but a gain, nothing but a complex gain. Hence, for non-linear systems also, we are going to associate a gain of this system, a gain of this operator and this gain of the operator we are going to call as eta. This is called eta, Greek alphabet for eta and this gain of system for input equal to A sin omega t. For this particular input, we are going to associate a gain of the system and this gain, we are going to allow it to vary as a function of omega, the frequency of the sinusoidal input. We can also allow it to depend on the amplitude of the input signal. So, what is the motivation for this? Because the motivation for linear time invariant systems behave like this. One can check that if G of s is a transfer function of a system of transfer function of an LTI system. Let us also assume that it is stable. Then, when you give input A sin omega t to the system G of s, what comes out is output equal to some transients. What is the meaning of transients? It goes to 0, stable, all poles in open left half complex plane. No poles on the imaginary axis also. Transients going to 0 plus steady state part. There is a steady state part that we are going to write now. This steady state part turns out to be nothing but A G of j omega sin omega t. This is the part we will focus on for the rest of this lecture and also the next lecture. When you give an input A sin omega t, this A sin omega t might have suddenly been switched on. That is why we might have some transients. But because all the poles of the transfer function are in the open left half complex plane, there is nothing on the imaginary axis, there is nothing in the right half complex plane. Hence, all these transients, because of the sudden switching on, go to 0. But eventually, there is a steady state part. This steady state part is, of course, it has sin omega t. This is what we like to think of as the forcing function, the forced part and the transients, the natural response and the forced response. We are speaking about the forced response. The forced response has exactly the same frequency sin omega t. If A was the amplitude, there is A here also. But there is a complex number associated to, complex number that is called multiplied to sin omega t. Of course, this is primarily notation. You give a signal that is real A sin omega t. How can the output become complex? I mean, how can it have an imaginary part? We measure signals that are functions of time, they are all real part. At any time instant, the value of function of t on the oscilloscope we measure, we do not measure complex signals. This is a notation. G of j omega is indeed a complex number and this complex number has some meaning. The real part and imaginary part have some important meaning, but the way it is written here, it is just notation. So, what is the meaning of this notation then, if it is just notation? So, if G of j omega is equal to some real part plus j times imaginary part. We can break its real part and imaginary part. There is some transfer function G of s. We evaluate it at a particular value of omega. When we evaluate G at some complex number, there is a very rare chance that it has become a real number. It could become a real number. For example, if the necklace plot for this omega intersects the real axis. Except for that rare case, this G of j omega has a real part and imaginary part both. Imaginary part by itself is some real number. It has got multiplied to j. Now, it has become purely imaginary and now you have added another real number. So, now together it has become a complex number. So, when we say G of j omega times sin omega t, what we mean by this is real part of G of j omega times sin omega t plus imaginary part of G of j omega times cos omega t. To associate the imaginary part with sin omega t and this j, this j when it applies to sin omega t, we will think of it as cos omega t. This is a very important fact. What is that important fact? You take a linear time invariant system with transfer function G. You give it A sin omega t. What comes out has a transient part that goes to 0 because G is stable. Stable by here we mean no poles on the imaginary axis, no poles in the right half plane and there is a forced part. This is also steady state part. At steady state, this A sin omega t will have an effect only on the forced part. The forced part is equal to real part of G of j omega times sin omega t plus imaginary part of G of j omega times cos omega t. So, one can check this by first principles. By substituting this into the differential equation, every transfer function after all is nothing but a differential equation between this and this. In the differential equation for output also, you substitute sin omega t and cos omega t. The coefficient that come out, I forgot to factor A here. If the amplitude is larger, the output also will have amplitude exactly larger, but how much part comes with sin omega t? How much part comes with cos omega t? That is determined by the real part and imaginary part of G of j omega. Now, you see there is no j here. Now, the output is a real signal. What happened to the imaginary part of G of j omega as far as our earlier notation is concerned? That has got translated to cos omega t amplitude. So, how do you, how does one verify this? We can take a very simple example and also verify this. Suppose, the transfer function of some system is 1 over s plus 1, input is A sin omega t, output is question mark. So, this particular, let us say input is, let us come back to the more traditional notation of using input U, output Y. So, the meaning of this transfer function is nothing but d by dt of Y plus Y equal to U. If this is the differential equation of a system, then we associate this transfer function to the system. Now, U is equal to A sin omega t. So, we will write U is equal to A sin omega t here. Now, for Y, of course, there is a e to the power minus t term, which is homogeneous for every initial condition there is a part, but that is going to die to 0. So, we are interested in the forced response part. So, we will write that this is some b times sin omega t plus c times cos omega t. We are going to find out values of b and c. So, we will substitute this into the differential equation here and when we substitute we get derivative of sin omega t is nothing but cos omega t. So, we get b omega cos omega t plus c cos omega t plus c omega sin omega t is a negative sign here, plus b sin omega t. So, this part here is the derivative of Y and this part here is nothing but Y. So, this is equal to A sin omega t. Now, sin omega t will equate to the coefficient will equate and cos omega t, there is nothing cos omega t here. So, sin omega t in cos, this in other words we say that sin omega t and cos omega t are independent functions of time. If this has to be satisfied for all values of time, then we can equate b omega plus c equal to 0 and A is equal to b minus c omega. So, we are going to solve these two equations simultaneously for obtaining values of b and c and by solving this we will get the values of b and c. We have two unknowns b and c and two equations in b and c. We will use this to find the values of b and c. So, what do this translate to? Let us write this into a matrix b c equal to one of them is of course A, other one is 0. So, the coefficients of b and c is 1 minus omega and the upper equation is omega 1. So, in order to get b c all we have to do is b c equal to inverse of this matrix applied to 0 A. So, the inverse of this matrix is obtained by just minus 1, minus 1 applied to 0 A divided by the determinant of this matrix. Determinant of this matrix is nothing but minus omega square minus 1. So, it is 1 over minus 1 over omega square plus 1. So, this is how we can get b and c. So, b and c is equal to this nothing but is second column of this. So, this is minus plus A over omega square plus 1 and here we have minus omega over omega square plus 1. This is the value of b and c. So, what we can do is we can check whether, so there is an A missing this time t. We can check whether these are indeed the real parts of G of j omega. So, G of j omega from the theory that I said, we have got output y of t equal to A times A times real part of G of j omega is the coefficient of sin omega t plus A times imaginary part of G of j omega times cos omega t. So, we will quickly check whether real part of G of j omega is nothing but 1 over omega square plus 1 and imaginary part of G of j omega is nothing but minus omega over omega square plus 1. After checking that, we would have ratified this formula that we wrote for linear time invariant systems, which is what we will use as motivation for defining the describing function of a non-linear system. So, G of j omega is equal to 1 over j omega plus 1. This is equal to 1 minus j omega divided by omega square plus 1. This is exactly equal to 1 over omega square plus 1 plus j times. So, this is what we also obtained here. We have obtained that the real part is nothing but B, B over A. Notice that there is a factor. The part that comes as sin omega t of course, has a factor A that comes because the input has A sin omega t. If you scale the input by a constant, of course, the outputs will get scaled by the same constant both for sin omega t and cos omega t. But how much portion has gone to sin omega t? How much portion has gone to cos omega t? For linear time invariant systems, that fraction does not depend on A, a very important point to note that linear time invariant systems, it is only the relative partitioning. Relative, I do not mean to say that they add up to 1, that is not the case. But A does not play a role in these constants, the real part of G of j omega and imaginary part of G of j omega. In other words, the magnification also has some contribution coming into cos omega t. But scaling by A does not change these relative contributions. So, real part of G of j omega indeed is equal to 1 over omega square plus 1, which is what we got as B, B except for this A factor. And the part coming with cos omega t was exactly minus omega over omega square plus 1, which is nothing but the imaginary part. This is what we will like to generalize to describing functions also, to for non-linear systems also using this notion of disrupting function, where we allow dependence on omega and also A for non-linear systems, that is very much possible quite expectedly. We will see that for the case of saturation non-linearity.