 Welcome to the 12th session and here we begin by answering the question that we raised in the previous session. Namely, if I have a continuous function at least a region in which the function is continuous, can I construct it by using more pulse? Let me now go back to the answer. So I have here this function which is smooth continuous in that region which I have identified and I have this very narrow pulse spanning an interval of delta and with a height of 1 by delta. You could visualize this pulse moving along the time axis, this function is a function of t whatever t might be, time for example and I move that pulse to bring it to a specific location. I have brought it to this location here, let us call it t equal to t0. So if this pulse were originally at 0 here and if I give this pulse a name, I call it delta s, capital delta t, this is the name I give to this pulse and this pulse which I have drawn here can be written as delta, capital delta t minus t0 shifted to lie at t0 instead of at 0 and the original function that I have here is x of t as a function of t. Now I multiply these two and let us see what I get, multiply them and integrate, integrate on all time. So what I am saying is take xt delta, capital delta t minus t0, integrate with respect to t for all t, all over the time axis. What do I get? Now here you need to do a little bit of thing. So visualize that you actually have this pulse lying at t0 as it were and essentially this capital delta is small enough that the function xt here is almost a constant on that interval and therefore all that happens when you multiply xt by delta delta t minus t0 is to capture or to pick that particular value of xt, maybe you could think of the value at the center, it does not matter. So in other words what I am saying formally in mathematical languages, formally xt times delta delta t minus t0 is almost equivalent to x of t0 delta t minus t0. Now you know you must appreciate what I am saying. The two sides of the equation here I am talking about this equation xt into delta t minus t0 is equal to or is equivalent to x of t0 times delta delta t minus t0. The two sides of the equation have a very different meaning. Here you have a product of two functions and here you have a multiple of the pulse. So what we are saying is in the region where the function is continuous, the product of xt with that narrow pulse is just a multiple of that narrow pulse. And now when I integrate it with respect to all time, I simply get that factor by which it is multiplied coming out because the pulse has unit area. Now you understand why we took unit area in the first place. So let us say it formally. What we are saying is integral xt times delta delta t minus t0 dt over all t is the same as integral minus infinity to plus infinity x of t0 times delta delta t minus t0 dt. But x of t0 is a constant, so it can come out of the integral and therefore you have that becoming equal to x of t0 times minus to plus infinity delta delta t minus t0 dt. And this is essentially the area in the pulse which we can conveniently make one or unity. Now what it means is the pulse, this is one way to understand it. The pulse is like a seam or it has a sifting property. In fact that is what that is the term used formally, the sifting property as they call it. What does that property say? It says x of t delta delta t minus t0 dt shifts out or pulls out x of t0. Now you could understand it just rightly different. You know we have also written an intermediate step there, let us write it again. We wrote down x of t delta delta t minus t0 is equivalent to x of t0 delta delta t minus t0. And then we said that when we integrate x t delta delta t minus t0 dt we get x of t0. Now let us take this particular pulse that we drew. Let us assume that it is symmetric about t equal to 0, the original pulse. Let us call it delta delta t and let us note that delta delta t is equal to delta delta minus t. If we assume that it is symmetric 0 as we have done here, use this property to write down and a slightly modified equation. The slightly modified equation is integral x t delta delta t minus t0 dt is the same as x t delta delta t0 minus t dt. And this becomes equal to x of t0. Now this equation has a totally different interpretation. It is beautiful how by tweaking the equations a little bit we come up with new interpretations. We will see the interpretation in the next session.