 A warm welcome to the 18th session in the 4th module of signals and systems. We have been working with rational Laplace transforms all the while in some of the previous sessions and we have been encountering one issue after another. In the last session, we encountered an issue where the numerator once you write down the rational Laplace transform as the ratio of two polynomials in S, you could have trouble with the numerator degree being greater than the denominator. Let us take an example. So, you could have something like S or S squared rather S squared plus S plus 1 divided by S plus 1. So, here the numerator degree is greater than the denominator degree, a bit of trouble. So, what do we do? We carry out a long division as we said. This is very simple. So, we have what is called a quotient, a polynomial in S and a remainder. Both Q S and R S are polynomials in S and now degree of R S is strictly less than the degree of the denominator. That is what we wanted. This long division comes to the rescue. Well, it comes to the rescue in being able to express this as a sum of two terms. But how do we deal with this term? What is this S? It is going to bring us to a new concept. So, we need to first ask what is the inverse Laplace transform of a constant? Let us see. Let us just take the constant C0. How would I get C0? In other words, what is the xt so that xt into e raised to the power minus st dt integrated on all t gives you just a constant C0. So, you would have to just pick something, one simple way is to just pick. And if you put xt equal to C0 times delta t, you get the answer. Indeed, integral C0 delta t times e raised to the power minus st dt is simply C0. So, therefore, the inverse Laplace transform for constant is very easy. Just an impulse. An impulse at 0 at t equal to 0 with strength equal to C0. Now, let us invoke the dual differentiation property. We call that we had a differentiation property where we differentiated the Laplace transform. But now we are talking about the differentiation of the time function, multiplication of a Laplace transform by S. So, what happens? So, you know C0 was the Laplace transform of C0 delta t. Of what would C0 S be the Laplace transform? It would require you to differentiate here. So, we brought in a new idea now. Can you differentiate an impulse? Already the impulse was troublesome, not even a function. So, the generalized function, it gave us some trouble in understanding. It took us quite a while to appreciate what this business was all about. It was in fact, what is generally called a distribution. It is a generalized function. How did we arrive at impulse? Let us see. So, where did impulse come from? You had this very narrow pulse, symmetric pulse about 0. And you took the so-called limit, though of course, limit is a bad word here. Limit as delta tends to 0. It is not a very correct word. But we did that. We essentially took the so-called limit and that gave us an impulse. Now, let us take the derivative of this pulse and allow for impulses in the derivative. You know, because the derivative does not exist in the true sense. There are discontinuities here. Suppose we take d dt of this. Let us draw it again. How would it look? It would have 2 impulses. One impulse of strength plus 1 by delta at minus delta by 2. I am expanding the situation. And the second impulse of strength minus 1 by delta at plus delta by 2. So, these are impulses. Now visualize capital delta going to 0. Two things happen. The impulse strengths tend to infinity. And the impulses come very close, arbitrarily close. It is called a doublet because there are two impulses and both of them are becoming infinitely strong and almost fusing. Two impulses becoming infinitely strong, becoming stronger and stronger and stronger and coming together and almost fusing. You know, in principle, what does this doublet do? If you think about it, suppose you applied this doublet upon a function xt. At a point, say t equal to t0, where xt is differentiable. So, let us visualize the situation. You have this xt. This is the point t equal to t0. And you put this doublet here. Let us put the doublet there. You have two impulses coming closer and closer to one another. This point is t0 minus delta by 2 and this point is t0 plus delta by 2. So, suppose you applied it. Apply means multiply by the doublet and integrate. So, let us write out the integral. What happens when you apply the doublet? Let us now analyze it algebraically. So, applying the doublet means multiply first and integrate. So, it amounts to saying you are doing this. So, do you see what we are doing? We essentially doing 1 by delta x of t0 minus delta by 2 minus 1 by delta x into x at t0 plus delta by 2. Let us see this on the graph that we had. So, essentially we are talking about picking these two points here. This one and this one. And we essentially subtract this minus this and divide by delta. So, it is of the form 1 by delta. It is emphasized that. So, essentially we are talking about a change in x between the two points divided by the interval between the points. And now we are asking that we take a limit of this. Now, what is the limit of this process? You have taken a change between two points very near one another. You are dividing by the interval and then taking a limit. It is nothing but the derivative. So, this is essentially the derivative. So, when you apply the unit doublet. Now, I have not said unit so far, but I am saying it now. You just said doublet all this while. When I apply the unit doublet upon x t at a point where it is differentiable, it essentially gives you the derivative of x t at that point. You know why it is called a doublet. You also know what the doublet does. Now, where does the unit business come from? The unit business comes from the fact that you got this as the limit of the derivative of what was trying to be a unit impulse, trying to be a unit impulse. Let me explain this to you again. So, what I mean by that is if you take this, this is trying to be a unit impulse as delta tends to 0. And when we take its derivative, we get a unit doublet as delta tends to 0. In contrast, if I were to multiply this, so if I have c 0 times 1 by delta here as the height, its derivative would give a doublet with strength. So, doublet can also have a strength, doublet with strength c 0. So, that means the process of just having an s or a power of s in the Laplace domain essentially is a differentiation process. So, you know this unit doublet is like a differentiation operator. It is just waiting to differentiate something. Again, it is not a function, it is a generalized function. So, in my expansion of a rational Laplace transform, the remainder part is no problem. For the remainder part divided by the denominator part, I can make an inversion by using the standard poly-x terms. But for this quotient part, I need to bring in the impulse and the doublet and then what more. So, if I have an s square term, I would need to differentiate the doublet again. How would I differentiate the doublet? I need to differentiate the doublet by differentiating two impulses. So, how would it look? Let us just get a feel. So, you have a unit doublet which is like this. So, it is of the form. Now, if I want to take a derivative of this, I would need to put two doublets. So, you would get 1 by delta times a unit doublet at minus delta by 2 minus 1 by delta times a unit doublet at delta by 2 which comes from here. So, together this would be the derivative. So, when I take a derivative for doublet, I get two doublets and those two doublets also start coming closer and closer to one another and now you have several heights going to infinity here. Now, all this is becoming difficult to visualize, I know. This is where we now if we really wish to understand these things, we need to go into the domain of what are called generalized functions or distributions. I shall not go any further in discussing these because they take us very much outside the scope of this course. But an awareness of these generalized functions or what are called singularity. These are all called singularity functions and they are not really functions to be very truthful. These so called singularity operators are essentially derivatives of the impulse. Their effect is easy to understand. When you put a unit doublet onto a function which is differentiable and integrate, it gives you the derivative of the function. When I put the derivative of a unit doublet, it gives you the second derivative and so on. So, essentially successive singularity functions give you higher order derivatives at the point where they occur. But these operators themselves are not very easy to visualize, but we have given some hint what they might be visualized to be. Well, we will see more in the next session. Thank you.