 A warm welcome to this discussion session. I think it is time we had a discussion with some of our teaching associates who would like to raise a few points and a few questions as there are some tricky concepts that we have been discussing over the last few sessions. Maybe I will go over to the teaching associates and they will tell you the questions and the points that they have. So, over to you. Hello sir, we had discussed about the concept of implicit and explicit functions. So, is it possible to relate it to a practical example that is implicit functions where are they used in the real life examples? Yeah, that is a good question. Now, let us look at the same example that we began with. If you look at the description there, the RC circuit for example, we took the output y of t which is the output voltage here and the input x of t which is the input voltage and we saw that the output is related to the input in the following way y t plus r times the current C d y t d t is the input. Now, here this is an implicit relation between the input and the output and in fact, many relationships are like this, they are implicit. One cannot read off, read off so to speak y t from x t. What I mean by that is that I cannot explicitly say well y t can be calculated from the x t's in the following way. I cannot put down a formula where y t is on one side of the equation and x t and I mean x t means x t at all time is on the other that is explicit. So, let me give you an example of an explicit relation y t is let us say d x t d t very clear. The output is the derivative of the input as clear as crystal. So, I can read off y t from x t time by time. I cannot do that in this RC circuit. I need to do some work. I need to solve it. Now, in fact, that is where this idea of the response of that RC circuit to a very narrow pulse comes into the picture. So, what I do is I say well if the RC circuit if you give a very narrow pulse, give a very narrow pulse, let the width of the pulse be delta or any multiple of delta for that matter and the same multiple of delta comes here one by delta. This is a very narrow pulse. Give this as the input to the RC circuit. So, it is a hypothetical voltage which you are giving to the RC circuit and you record the output. Let that output be H of t. The beauty is we will show later that we can write y t explicitly in terms of x t and h t provided delta goes towards 0. So, in a way what we are going to do in the next few sessions is to make this evident to ourselves that many real life problems give you implicit relationships, but you want the relationship to be explicit. So, you need to do some work once and that work suffices to make that relationship explicit for you for all possible x t. We are going to talk about the response to an impulse. I hope that answers your question. Do you have some other observations? We discussed in the lectures the shifting property. I think I have a way to illustrate it. So, can I just show it to you and you can comment on it whether it is good or bad? Very good. That is very nice to hear. So, in fact the shifting property is a very important idea in the context of an impulse and Sushrut says that he has a much better way to illustrate this or a clear illustrative example. So, let me now invite Sushrut to put his example before you and let us see how he does in terms of illustrating this in a much better way. Yes, welcome Sushrut. Please come. So, here Sushrut before me and he is now going to illustrate the idea of shifting in a much better way. Yes, Sushrut. Why do not you tell us what you are going to do? So, what am I going to do is I will take an example which is e raised to minus t and I will take an impulse and I will go through the process of how we reached at the conclusion that the value of the function at the point where the impulse lies comes out. He is going to illustrate the shifting property to you with a reasonably good example and you know actually going through the math carefully. Yes. So, here we have a function e raised to minus t. So, this is one and this is the function somewhat and then we have this impulse. I am choosing it at one we can choose at anywhere else. I am just keeping it its with delta right now. So, the height will be 1 by delta and then now we need to multiply these two. So, after multiplication the part outside the delta function will be reduced to 0 because delta function is only finite value between 1 minus delta by 2 and 1 plus delta by 2. In fact, Sushrut let us illustrate that with a color. So, let me help you with that. So, you could actually take let us say put this pulse here. So, this is t equal to 1 here, is not it? So, you could put the pulse here, let us put it like this and you are centering the pulse at 1. That means this point would be 1 minus delta by 2 and this point would be 1 plus delta by 2 and the height would be 1 by delta. Now, you put it on the same one and what you are saying is that what is outside the red box goes away. So, perhaps we could illustrate that below you know. So, you could just say you have this left here teeny-weeny piece like that. Yes. Now, you can go on. So, we see that the there is some difference of values between the two extremities. So, we will just go ahead with that and then we will make the pulse narrower as we go on. So, we will just. So, we will just multiply them. So, let us just calculate the integral. Yes. Let us just calculate the integral. Let us just calculate the integral. So, you raise the power minus t multiplied by this over all time. So, then that boils down to 1 minus delta by 2. The value of delta is 1 by delta between these two points. That is very good. So, we can go ahead like this and delta is a constant with respect to t. That is. So, it comes out. So, it will be minus e raise to minus t with the limits 1 plus delta by 2 and 1 minus delta by 2. So, it was 1 by delta times e raise to minus 1 minus delta by 2 minus e raise to minus 1 plus delta by 2. That is correct. I think that is it. So, we can again draw the figure. It was looking something like this. So, when we make the pulse narrower and narrower, the height difference between the two extremities will be reducing and it will finally be something like this. Almost. So, we can see that y component of this pulse is almost constant. And if we put that in the equation we have, that is take limit delta ending to 0. We will find the area of this function. So, let us do that. And we can take e raise to minus 1 common from both of these. And it will also come out of the limit because it is not dependent on capital D delta. Very good. Now, you can tell them how you calculate this limit. So, it is of the form 0 by 0 if we directly substitute delta equal to 0. So, maybe you can use Lopital's rule and take derivative on both sides and the limit will not change. So, it will be and now we can directly substitute delta equal to 0. Very good. So, Lopital's rule gives you a very convenient expression. Yes. It is e raise to minus 1 or 1 by e. Very good. So, that is how you can go through the process. And because we derived graphically, it is much more intuitive than just the mathematical part. So, in fact, here it clearly illustrates to you with a simple example how this impulse shifts out. You know, the word shifting actually suggests what is done in the word shifting means to identify small particles. You know, when you shift out, shifting in grains, for example, people shift grains. So, they identify small little contents in the grain. And shifting here really means identifying that particular value of xt at the point where the impulse lies, bringing that out, taking that out of all the sand lying around it, so to speak. Very nice. So, Shrut, very good. That is a very nice illustration. I invite all the participants of this course to construct such examples for themselves to understand the shifting property better. This idea of impulse and the way impulses come together to construct a continuous function needs to be understood very well to appreciate the subsequent discussions of the course. Thank you, Shrut. Thank you, sir. So, do we have any other questions on the topics that we have talked about so far? Sir, you just explained about the direct delta function. And it seems to be a very important concept in signals and systems. I just wanted to ask if it is also very important concept in some other fields of physics or probably mathematics. For example, I have heard that it may be used in, say, discrete probability densities or something like that. Could you just elaborate on that? Yes, that is a very important question. In fact, in a way, the dark delta as we are going to talk about, the impulse, is one way to bring in a connection between what you call continuous variable, continuous random variable densities and discrete random variable densities. Let me illustrate to you with an example. So, for example, many of you are familiar with what we call the Gaussian density, you know, for the normal distribution. Look something like this. There is a mean here. And of course, the expression, if the mean is at t equal to t0 and t is the random variable, then it looks like e raised to the power minus t minus t0 squared by 2 sigma squared divided by an appropriate constant. Let us not get into that for the moment. The constant depends on sigma. Now, this is of course, what you call Pt, you know, so now you could call t the random variable and Pt at a particular value t. This is the probability density function of the random variable t. Of course, you know, to be very careful, I should use a different name for the random variable and the value. So, maybe I will write t1 here if you really want to do that. But that is besides the point. The point is, if I take a particular value t equal to t1 and ask, what is the probability that random variable will take on that particular value? The answer is always 0. So, what is the probability that this Gaussian or the normal random variable will take on a value t equal to 5? Although the density is non-zero at 5, the value of the probability is always 0. On the other hand, if you ask me, what is the probability that that random variable would take a value between 5 and 5.5, I can calculate. I can integrate the area under the probability density and that gives me the probability. And of course, that area has to be less than 1 because this density integrates to 1. Now, in contrast, suppose I am talking about a fair die, right. So, one fair die, so which has 6 faces, 1 to 6, let us say. What is the probability that the die shows 3? It is 1 by 6, of course. How do we represent this on the random variable space? So, you have the random variable here equal to the value shown by the die. And that can take discrete values 1, 2, 3, 4, 5, 6. How do I write a density here? I would write a density here by putting impulses at each of these places. I would put a derap delta here. Each impulse would have a strength of 1 by 6. So, I need to bring in, you know, now I can ask what is the probability that the die takes on one particular value because you can integrate. The impulse captures an area, a non-zero area into 0 length. So, what is the area contained at the point 3? The answer is 1 by 6. So, the idea of the derap delta is very important in going from continuous densities to discrete densities. That is the answer to your question. Thank you.