 So, welcome to the 17th discussion that I am going to have with you. In the previous discussion we had seen a very beautiful relationship between the unit step and the unit impulse. The unit step was the running integral of the unit impulse and consequently it should be true that if I take the derivative of a unit step it should give me the unit impulse, but we saw a problem there. The unit step is not a continuous function. So, actually it is inappropriate if you wish to call it that to take a derivative in the sense of traditional calculus, but here is where we are going to start to deviate a little bit in this course on signals and systems. We are going to accept that we will allow derivatives of discontinuous functions, but of course those discontinuities must be isolated like they are at the unit step. The unit step has only one discontinuity. So, let us draw it. So, 0 all the way from minus infinity up to this point and 1 from this point onwards that is the unit step and we have agreed that this is discontinuous at an isolated point. So, discontinuity isolated. So, we are saying that we shall accept its derivative and now just to complete what we were trying to do last time, we shall actually find the derivative of this in the traditional way that we do in calculus. How do you find the derivative? You would shift the function a little, subtract the shifted function from the original function, divide that by the shift that you applied and take the limit as the shift tends to 0. Let me write that down. What I am saying is if I want to find dx t dt, I am essentially going to take x t plus say a little tiny delta minus x t divided by delta with a limit as delta tends to 0. Now of course this should also be equal to limit as delta tends to 0 of x t minus x t minus delta divided by delta and for that matter also the limit as delta tends to 0 of x t plus delta by 2 minus x t minus delta by 2 divided by delta. Now all this is fine when we are talking about a function which is continuous at a point t. There is no problem. In fact continuous and differentiable. So, if the right limit of the derivative and the left limit of the derivative is the same then all this is true. So, in traditional calculus we do not accept derivatives at points where the function is discontinuous. In the unit step therefore we will have trouble at the point of discontinuity. In principle you could take a derivative everywhere else, but not quite at the point of discontinuity. But now in this course we are going to bring in these generalized functions as we have agreed and therefore that notion of derivative is acceptable. So, therefore let us take the first definition that we have written down here for the derivative namely shift it forward subtract the original function from the shifted function and then divide by the shift with the shift tending towards 0. So, let me illustrate what I am saying by referring to the specific expression that I have written here. You know what I am saying is that here unlike what you might understand in traditional calculus where you say you are looking at the change in the function as it goes away from t, we are interpreting this in the language of signals and systems and calling this a shifted version. So, a shifted version of x t and this is the shift is a slightly different interpretation of the derivative where the expression has not changed, but it is more signals and systems like if you know what I mean. The reason why I am emphasizing this so much is I am going to use the properties of linear shift invariant systems now. So, I am going to ask a general question about linear shift invariant system which will help me answer what is the unit impulse response of that RC circuit. To answer that question it is not easy to apply an impulse we do not even understand what applying an impulse means at first go to the circuit. So, it is of course not too difficult to understand what happens when we apply a unit step. Let us complete that discussion anyway. We know what this RC circuit does when you apply a unit step. When you apply a unit step to this let us denote the unit step by ut. So, ut is this function what happens this is the input and let us draw the output on the same graph. We assume the capacitor is initially uncharged. So, all the while when the input is 0 nothing happens the capacitor remains uncharged and then afterwards we realize that the capacitor cannot have a sudden change of voltage across it. So, all the voltage would appear across the resistor initially and therefore, the capacitor voltage would rise steadily from 0 steadily with an exponential pattern. So, it would look like a asymptotic exponential rise and we know the expression. So, we will call it the unit step response here and let us denote the unit step response by S of t. S of t as this expression as this figure shows is essentially 1 minus e raised to the power minus t by tau times ut where tau is the product of resistance and capacitance. Now, you can check this expression you know as at t equal to 0 S t is equal to 0 as t tends to plus infinity S t tends towards 1. So, this fits our whole picture. Now, let us ask a general question. I know that if I took the derivative of a unit step I should get the unit impulse, but then I need to justify that by the definition of a derivative and let me do that now. So, let us ask what is the derivative of a unit step in the traditional sense. So, it should be something if we take the first definition it should be something like ut plus delta minus ut divided by delta with a limit as delta tends to 0 and let us sketch ut plus delta. So, ut plus delta starts from minus delta here and let us sketch ut of course, ut starts delta after this is ut and this is ut plus delta and remember you are subtracting the red function from the black function and you are dividing this difference by delta or multiplying it by 1 by delta. Let us draw that difference and scaled version. So, what we have is essentially this. So, ut plus delta minus ut into 1 by delta as it where will look like this from minus delta to 0 and a height of 1 by delta and there again you can see that this is the famous narrow pulse that we have been discussing all this while narrow pulse going to an impulse as delta tends to 0. So, we are in good shape we have verified even by using the notion of a derivative as we understand in traditional calculus that if I took the derivative of a unit step I would get unit impulse. Now, I leave it to you to verify this by using the other two expressions. So, this is what we call a forward difference you could take a backward difference namely ut minus ut minus delta divided by delta or you could take a balanced difference ut plus delta by 2 minus ut minus delta by 2 divided by delta all of them would really give you a very similar looking narrow pulse which would move towards an impulse as the width tends to 0. Now, the general question that we need to ask is what happens when I take the derivative of the input to a linear shift invariant system? What happens to the output? So, let me frame the question and we shall answer that question in the next discussion after invoking the appropriate principles of linearity and shift invariance. Let me frame the question. So, we know what to expect in the discussion that will come. The question is I have this linear shift invariant system here. Let us call it S and I give the input x t to it and get the output y t. What would happen? What would the output be if x t is replaced by dx t dt? What would happen to y t? That is the question that we want to answer and I am sure you must have already thought of an answer, but we will formally answer it in the discussion that will ends you very shortly. However, do think about it before you come back to the next discussion. Thank you.