 So, welcome to this session, where we now take further the very important principle that we derived in the previous session, namely the impulse response of a linear shift invariant system and its importance in the characterization of that system. We would do well at this stage to consider an example, but in order that we understand that example, we need to build an auxiliary principle, which we shall do in this session after first explaining the example that we want to consider. Let us go back to that famous simple first order system that we saw, which inspired us to build an analog from a mechanical and an electrical system and we shall take the electrical system as the exemplar here or the context in which we want to discuss the example. So, let us go back to the simple RC circuit. So, we have this resistance and capacitance excited by an input voltage and studied in terms of the output voltage. Now, we have analyzed this system before. So, we do not need to go through the whole analysis again, but then we do understand here an important principle namely that this system is linear and shift invariant. So, if we accept the result of the previous discussion that we have just completed, then we should be able to characterize this system completely by means of its impulse response. So, we would like to ask what is its impulse response? This should give us an interesting answer and what does that mean? You see if I wish to find the impulse response of this circuit, theoretically I should be able to apply a unit impulse at the input and seek the output. What do you mean by unit impulse? A unit impulse voltage, remember. So, let us put that principle apply a unit impulse. Now, what does this mean? You see, what is a unit impulse of voltage? Let us go back to the definition of an impulse. So, we said an impulse is the limiting situation when I have this tiny little pulse with a width of delta and a height of 1 by delta. Thus maintaining the area at 1, 1 by delta times delta which is 1 even as delta tends to 0. Now, what is the area of a voltage? You know what physical significance does it have? Now, you know for this we will have to look at the equation that describes the capacitor or the equation that describes anything that has a derivative operator embedded. Now, the current in a capacitor is related to the derivative of the voltage. Is that right? So, let us see. So, we have for a capacitor the current is well, if the capacitance value is c, then the current is c d voltage dt. And if I integrate both sides, then I have the integral of the current with respect to t is c in principle v. So, in some sense, there is an integration operation involved on the equation of a capacitor. Now, we will understand this better when we actually complete the derivation. Just let us keep this idea in mind right now. But what it tells us is that instead of trying to deal directly with applying an impulse and then trying to find the output, we should do something indirect. We should try to deal with an integrated form, you know because then you have a voltage, deal with an integrated form and then come to the conclusion of the impulse response. Sounds a little mysterious, but I will explain. So, let us ask the question, what would happen if I created an integral of the impulse? What would I get? Right? So, let us put that down. What if I integrated an impulse? Because, you know, the integral of an impulse gives us something interesting. Now, let me again do this by using a limiting argument. So, I have this tiny little pulse here of width delta and height 1 by delta. And this point, let us call it t0. So, I create, this is the variable t and I have delta, delta located suitably. So, you know, I will say delta, delta t0 just to make it clear that we are locating this at some place of t. Now, delta, delta t0, t is the function I have drawn, not quite an impulse, but moving in that direction as delta tends to 0. And now, I consider the integral of this from minus infinity to every t. This is called a running integral. We call it a running integral of this impulse or this very narrow pulse which is moving towards an impulse. And let us see what I would get. Now, you know, there are three situations possible. Let us draw all the three situations. So, I identify this point t0. This point is t0 plus delta. This height is 1 by delta. And this variable is t. And now, I identify three regions, region 1 up to here, t less than equal to t0, region 2. So, you have t between t0 and t0 plus delta. And finally, region 3, where t is greater than or you can even say greater than equal to t0 plus delta. And we will take these three regions separately. Now, you can visualize the running integral. The running integral captures all the area that has come up to that point t from the beginning of time. So, I will straight away, it is not very difficult to calculate this running integral. You just need to look at the total area that we have encompassed up to that point t. And you can see that in region 1, any way that area is 0, there is nothing to encompass. It is only in region 2 and in region 3 that we have encompassed some area. Let us look at it. So, in region 2, the area encompassed grows linearly. And finally, in region 3, all the area, all the unit area has been encompassed. And therefore, we can now draw the running integral. So, the running integral is minus infinity to t delta delta t0 lambda d lambda and can be drawn as follows. Up to t0, that integral is 0. From t0 to t0 plus delta, the integral rises linearly and then remains constant at 1. This part is linear. Region 1, region 2 and region 3. And now that we have drawn this, it is obvious what is going to happen as we start reducing that width towards 0. The running integral is going to go to a unit step. So, we will become a unit step at t0 as delta tends to 0. So, I have made a very important conclusion in this little discussion that we have had. The running integral of a unit impulse is a unit step located at the same point where the unit impulse lies. Now, what is a natural corollary? Let us take the derivative of both sides. So, unit impulse running integral gives you a unit step at the same place. So, if I want to go the other way, I should take a derivative. You see, what is the derivative? When you take any x lambda d lambda and make a running integral on it, let us call it x i of t. What is d x i t d t? i stands for running integral. What is d x i d d t? It is very clearly seen to be x of t. That is not too difficult to do. One can easily derive it from basic principles of calculus by making the perturbation on t. And in fact, I leave it to you as an exercise to complete this little proof that the derivative of a running integral gives you back to integrand. So, what is the conclusion we have here? Before we go to the next discussion, the derivative of a unit step, whatever that might mean, you see, it is not fair to talk of the derivative of a unit step actually, because the unit step is a discontinuous function. But then now I am being able to generalize this load. So, in the generalized sense, the derivative of a unit step is the unit impulse. Now, with this very important conclusion, we shall proceed to our next discussion where we will use this idea with greater depth and meaning. Thank you.