 So, welcome to the 13th session on the course on signals and systems. Let us take a quick moment to recapitulate the important result that we saw in the previous session. It was obtained by a small rearrangement of the shifting property. So, let us recapitulate that idea. We had said that an integral from minus infinity to plus infinity x t delta t minus t 0 gave us x at t 0. And we noted that we could choose delta t symmetric about 0. So, we could choose delta t like this. I am showing delta as a function of another variable which we call lambda. And here of course I must clearly make it a delta s capital delta to emphasize the point that it is spread over an interval of delta. So, delta delta lambda is equal to delta capital delta of minus lambda. So, it is an even function of lambda. And therefore, we said that delta delta t minus t 0 is equal to delta delta t 0 minus t. And therefore, we came to the important conclusion that integral from minus to plus infinity x t delta delta t 0 minus t. Of course, here the integral was with respect to t and here that is also with respect to t is equal to x at t 0. If you have this integral should have been with respect to t that is all right that is understood. But now you see there is something important that we should note here. You see the meaning of this equation here, this equation and this equation here have a totally different meaning. The first equation says that when I multiply x t with delta delta t minus t. So, when I multiply a continuous function with a narrow pulse and integrate it keeping that pulse of unit area I get essentially pick up or shift the value of the input or the function x t at the point at which the pulse is located. Look at the second equation. The second equation says something totally different. It says if you take t 0 now to be the variable. So, we will now think of t 0 as the basic variable by which the function is indexed. And this is the variable of integration t here. This is the variable of integration. So, what we are saying is now take any particular point t. t 0 is the index variable on the input. So, what we are saying is if you take the value of the function x at a particular point t which is x at t and multiply it by a narrow pulse which is essentially centered around any point t 0 and take a sum over all such t and that sum becomes an integral in the limit. A totally different interpretation. Listen carefully. I am repeating it. Let us look at the equation again. Any particular point t, the value of the input at that point t and impulse or a pulse for the moment not an impulse, a pulse placed at the point t with the indexing variable t 0. What I mean by the indexing variable is the variable with which with respect to which you describe the function x, integrated over all t. Now, the integral here should be interpreted as a limit of a sum. So, integral is essentially a limit of a sum. So, you know you could think of it like this. You could think of it as x at some t 1 delta delta t 0 minus t 1 plus x at some t 2 delta delta t 0 minus t 2 plus and so on. Take these t 1 t 2 t 3 and so on as close as you can make them. In fact, what we are saying is let me show it graphically. You know you must understand the same thing in many different ways. So, let us show it graphically. What we are saying is pick a t 1 here, pick a t 2 there, pick a t 3 there and this variable is t 0. That is what I mean by the index variable. Put a pulse here, put a pulse there, put a pulse here and this width as you know is delta. That is why we are calling it delta capital delta. This height would now be 1 by delta multiplied by x at t 1. Now, do the same thing at t 2. Let us use a different color there. So, again you have a width of delta here and a height of 1 by delta times x at t 2 here. And once again put another pulse here with delta height 1 by delta times x at t and so on. Now, visualize putting pulses at every t 0. So, you know here you have x at t 1, you have x at t 2 and so on. And maybe you have a function that goes let me show that function in some let us say the red color. You have some continuous function here which goes like this perhaps. This is actually you know graphically we are trying to show 1 by delta times x of t 0 here. Visualize putting pulses at every t 0. And now because you have a continuum of pulses you see there are not discrete points, but there is a continuum of pulses at every point there is a pulse so to speak. And all these pulses need to be combined and their combination leads you into the function x at t 0 under the integral sign. So, you see the integration is required because you have unit area contained in each pulse. So, go back to the diagram. So, you visualize putting pulses at every t 0. And now one more thing you need to do is to visualize delta going towards 0. I know this requires a little bit of work you know you do need to visualize first an infinity a continuum of pulses and then you need to visualize each of those pulses becoming smaller and smaller in widths and all coming together under integration. And then what we are saying in this equation let us write that equation again. What we are saying in this equation namely x at t 0 is equal to now here I intentionally show it in a different color x will show the variable of integration in a different color. So, x of t and then delta delta. So, here t 0 I will keep as it is and then minus we have a variable of integration again. So, here this makes a totally different statement. This says that every function is a combination of an infinite number of such pulses such very narrow pulses. This is a very important idea. Now these very narrow pulses that we are talking about here. So, these very narrow pulses with a height of 1 by delta and and a width of delta which means they enclose an area of 1 area enclosed with delta into 1 by delta which is 1. The area remains intact even as the pulse becomes narrower that is delta tends to 0. This ever narrowing pulse is a new idea to us and we shall talk about it more in the session to come. Thank you.