 A warm welcome to the 38th session in the first module of the course Signors and Systems. We have been looking at convolution in the last few sessions and we have been trying to understand how to carry out this basic operation between two signals that emanates from basic considerations in the context of signals and systems. In this session, we would like to embark on a study of two very important properties of this operation, convolution. And though I had hinted at those properties earlier, in fact, I have kind of used them, I would now present a formal proof of the two properties. And the two properties that I am going to discuss are as follows, two important properties, commutativity and associative of convolution. What we are going to do, we are going to prove these two properties formally and together for discrete and continuous convolution. I had given you an exercise, you will recall much earlier on in our discussions, I had given you an exercise to prove these properties, but now I will complete that task for you both in continuous and in discrete independent value. So, let us take up the first somewhat simpler property first, namely commutativity. Essentially says that x1 convolved with x2 is the same as x2 convolved with x1 and this is true both in continuous and in discrete time, true in continuous independent variable and in discrete independent variable. We shall prove it for both, let us prove it first for discrete, that is a little easier. Let y be equal to x1 convolved with x2, of course x1, x2, y are all sequences, so yn is by definition x1k, x2n minus k summed over all k, running over all the integer. And now we make a very simple transformation of variable, we put n minus k equal to l, remember this is true for each n and for that n, we put n minus k equal to l, which implies n minus l equal to k and when k goes from minus to plus infinity, then l also goes similar and therefore we could replace in the sum n, we could replace k by n minus l and we could replace n minus l minus n minus k by l and there we are, it tells us that y of n is also summation l going from minus to plus infinity x2l, x1n minus l, because essentially x2 convolved with x1, for every such n, x2 convolved with x1 evaluated at n and of course this is true for every such n, so naturally the convolution operation is commutative, so we have proved it. Now let us take the continuous independent variable. Here again let us write y is x1 convolved with x2, the only thing is x1, x2 and y are all continuous variables, which then goes on to say that y of t is integral x1 lambda x to t minus lambda d lambda running from minus to plus infinity. And we will take a q from the discrete case and we will make a similar transformation of variable, put t minus lambda is equal to alpha, where upon lambda is t minus alpha and when lambda runs from minus infinity to plus infinity, alpha run, now note it runs from plus infinity to minus infinity for a fixed t and t lambda is minus t alpha, so all in all we must now make the replacement of variable as follows, let us do it here. t minus alpha needs to be replaced by lambda, well t minus alpha by lambda or t minus lambda by alpha it is the same thing. And d lambda is minus t alpha and here this would go from plus infinity to minus infinity, so what we see in red is the modified form of thing and now we need to put the integral back in the form which is familiar to us, so let us write down integral first, y of t is plus infinity to minus infinity x1 t minus alpha x2 alpha minus d alpha, but then now you can take the minus sign and absorb with the reversal of integral which makes it minus to plus infinity x2 alpha x1 t minus alpha d alpha and that is the same as x2 convolved with x1 evaluated at that specific t, this is true for every t, hence we have proved that y is in general the same as x2 convolved with x1, so we have proved what we wanted to. So, we proved the commutativity of convolution both in discrete independent variable and in continuous independent variable, now we shall take up the question of associativity. We will first do it in the discrete independent variable case as before and then go to the continuous independent variable. So, let us take the meaning of associativity. Associativity as the name suggests is a question of association between pairs, so what you really want is to consider x1 convolved with x2 first and then with x3 and to question whether y is also equal to x1 convolved with the result of convolution of x2 and x3. So, if convolution is associative, yes, let us prove, let us begin with the discrete case of electronics, for it x1, x2, x3 and y all be sequences, the discrete case. Whereupon, we first consider x1 convolved with x2 evaluated at n and that is summation k going from minus to plus infinity x1 k x2 n minus. Now, x1 convolved with x2 and then convolved with x3, all this evaluated at n is essentially summation l going from minus to plus infinity, x1 convolved with x2 evaluated at l multiplied by x3 evaluated at n minus l and we can expand. This is the expansion and now we need to make a change of variable. But to make that change of variable, we must first see where we want to go. So, let us write down the result of the other side, x1 convolved with the result of convolution of x2 and x3. Let us write down the expression there. So, look at this expression here. Here, we have essentially calculated x1 first convolved with x2 and then convolved with x3. Now, what we shall calculate is x1 convolved with the result of convolution of x2 and x3. Now, first let us convolve x2 and x3 and evaluated at n. That is very simple. Summation k going from minus to plus infinity x2 k x3 n minus. And of course, so as not to confuse it with the previous k that we had, let us call this variable k1. Now, what we want to do is to calculate x1 convolved with this result at the point n. And that is going to be equal to summation l going from minus to plus infinity or if you like l1 going from minus to plus infinity x1 l1 x2 convolved with x3 evaluated at n minus l1 and we can expand this. This is what we have. Now, we need to make a correspondence of variables and then decide upon an appropriate transformation. So, let us go back to the previous expression that we had. Look at this expression here. You see, what you had here was this expression finally and you need to bring a correspondence with this expression here. Now, what is the correspondence? It is very clear that x1 has essentially the same kind of variable. So, there is a correspondence between l1 and k. I will show you both of them again. With the correspondence between l1 and k and we need to make a correspondence between l minus k and k1. Look at it, l minus k here and k1 there, k1 here, l minus k there. So, let us make that correspondence. Now, essentially what we are going to do in proving associativity is to make this correspondence. And for the discrete case, there is no problem. We will have no difficulty in making the correspondence and then transforming the double summation in such a way that we make the summation look the same. But in continuous variable, we will have to do a little work in the sense, we will have to bring in the notion of a Jacobian. And we shall do that in the next session, both for the discrete case to complete the argument and for the continuous case to add those few little details which will make the proof complete. We will meet again in the next session and complete the discussion on associativity. Thank you.