 So, welcome to the 18th discussion that we are going to have together and here we are going to answer the question that we posed at the end of the previous discussion, namely what happens when I take the derivative of the input applied to a linear shift invariant system. As I said, we are going to use the notion of a derivative that we know from calculus but with a slightly different interpretation, more signals and systems like if you please. And that will help us answer the question very easily. So, let us get down to business right away. So, I have this linear shift invariant system that we call it S and I have given it an input x t and it produces an output y t. Now, let me do this step by step. So, I will take the same S and give it x t plus delta and I shall invoke the shift invariance of the system and conclude that the output would be y t plus delta. Now, let me invoke additivity and homogeneity. So, first let me invoke homogeneity. So, I have applying minus x t to the same system S should produce minus y t and then additivity, this is homogeneity for you homogeneity with a scaling of minus 1 and we now invoke additivity. So, x of t plus delta minus x t should produce y of t plus delta minus y t and I once again invoke homogeneity now by multiplying by 1 by delta homogeneity again with a scaling of 1 by delta. So, I have x of t plus delta minus x t divided by delta or multiplied by 1 by delta applied to the system should produce 1 by delta times y of t plus delta minus y of t and now I have a very simple job. I just take the limit as delta tends to 0 on both sides assuming that that limit operation makes sense that means I have a limit on both sides to make delta smaller and smaller and then the conclusion is very easy. In fact, limit as delta tends to 0 gives d x t d t here and limit as delta tends to 0 gives d y t d t here. So, there we are this is a very important very simple principle actually. It says taking the derivative at the input of a linear shift invariant system results in the output also being subjected to a derivative operation. Now, we are all set to find out the unit impulse response of the RST circuit that we had put down in the discussions that preceded this. So, let us do that. The unit step response as we had it here for this RC circuit let me you know kind of show this RC circuit inside that block linear shift invariant system. We had agreed the RC circuit is an example of a linear shift invariant system. We gave it the unit step and we know the unit step response. The unit step was ut and the unit step response st as we called it was 1 minus e raised the power minus t by tau times ut where tau is equal to RC r is the value of the resistance in ohms and c is the value of the capacitance in Farad's. Now, applying the derivative principle take the derivative here and that should give me delta t a unit impulse. That should result in also taking the derivative here and now let us take the derivative. What is the derivative of st? In fact, dst dt as it were can be found by the chain rule or the product rule. So, it is dt of 1 minus e raised to the power minus t by tau multiplied by ut plus 1 minus e raised to the power minus t by tau multiplied by dut dt. But we know what this is. This is the unit impulse and this is easy to find. It is essentially minus minus 1 by tau times e raised to the power minus t by tau. So, there we have a very simple conclusion that we are drawing. dd t st which is also the unit impulse response is essentially 1 by tau e raised to the power minus t by tau times ut plus 1 minus e raised to the power minus t by tau times delta t. Now, let us look at this expression carefully. This expression essentially puts an impulse multiplied by an expression. Now, when you multiply a continuous expression by an impulse, you need only see what that expression assumes as a value at the point where the impulse flies. Everything else about that expression is irrelevant. So, going back to this expression here, this is the same thing as 1 minus e raised to the power minus 0 by tau delta t. And this is essentially 0. That is because this is 0. So, 0 delta t is essentially 0. So, that you know expression multiplied by delta t is not required. There is really no impulse there. Impulse has been scaled by 0. So, it is destroyed. So, we have a very simple and beautiful expression for the unit impulse response of an RC circuit. And let us write that down. That is the unit impulse response of the RC circuit is 1 by tau e raised to the power minus t by tau ut. Let us sketch it. It looks like this. There is a sudden jump and then there is a drop. And how much is this height here? This height is 1 by tau. You know, we need to understand this. What is this 1 by tau height here? We are talking about a unit impulse response to a unit impulse voltage, you know. So, what is a unit impulse voltage? There is a unit area contained in that voltage. And I am applying this unit area all at once at an instant. And you have a sudden. So, what this sketch shows is that the capacitor would record a sudden change of voltage. That seems to be rather contrary to what a capacitor normally does. That is why I had said we should postpone the discussion of what the unit impulse response is until we have established some of these corollaries or auxiliary principles. Because you know, this is somewhat counter intuitive signal systems is for good reasons and for the betterment of us going to be counter intuitive at points. A capacitor suddenly undergoing a change of voltage. How on earth can that happen? That can happen only when there is a sudden amount of charge pumped into the capacitor. And you know, think about it. You applied a unit impulse voltage to the RC circuit. That means, when there is a unit impulse to the sudden change of voltage in principle, that cannot go to the capacitor. It must go to the resistance. That means, the resistance must see an impulsive current. And that impulsive current must of course, go to the capacitor 2. And when there is an impulsive current, it must integrate into a finite amount of charge. And that finite amount of charge is what suddenly puts a voltage on the capacitor. How on earth can you get a jump of voltage on the capacitor? Only by suddenly pumping in charge. And that is what you are doing. You are putting an impulse of current onto the resistance which has no choice but to go to the capacitance. And once it goes to capacitance, it puts in. You see, what is an impulse? An impulse holds finite area. So, an impulse of current means a finite amount of charge. That charge has got pumped into the capacitance suddenly. And it has suddenly elevated the voltage of the capacitance. And we know by how much? You see, how much is the current in the resistance? It is the voltage divided by the resistance value. And that current integrated captures a unit area of that means, you know, the area of that current is 1. That means, the charge is 1. A charge of 1 divided by the resistance, remember, current integrated over time gives you the charge. Current is delta T by R. So, charge is 1 by R. And 1 by R divided by C gives you the voltage. That indeed is 1 by T. We will come back to this in the next discussion. Thank you.