 Welcome to the 39th session in the first module of the course Signals and Systems. We had embarked upon a discussion of commutativity and associativity in the previous session and in this session we intend to complete our discussion on what was left namely associativity. Now we had brought the proof that convolution was associative up to the following point. Let me enunciate what we got on both sides, the left hand side and the right hand side. Let me put it that. So, we continue the proof, the proof of associativity of convolution and essentially what associativity says is x 1 convolve with x 2 and then convolve with x is the same as x 1 convolve with the result of convolution of x 2 and x. We call this the left hand side and we call this the right hand side abbreviated by LHS and RHS. Now the left side evaluated at the point n essentially summation k and summation l and the right hand side again evaluated at the same point is summation k 1 and summation l 1. Now we need to make a correspondence and that is what we are going to do. So this is the correspondence between k and l 1 is clear and we also need to make a correspondence between l minus k and k 1. So let us do that, l 1 corresponding to k and k 1 corresponding to l minus k. So of course, n minus l 1 minus k 1 would correspond then to n minus k minus l minus k which is n minus l and disagrees with the left hand side expression. Moreover, when k goes k and l in fact both independently go from minus to plus infinity as you can see here you know it take k first. So you have k going from minus to plus infinity. If k goes from minus to plus infinity then of course, l 1 goes from minus to plus infinity and now fix k with a fixed k when for every k I mean you know for every fixed k when l goes from minus to plus infinity then so does k 1, k 1 also goes from minus to plus infinity. So therefore, k l independently going from minus to plus infinity implies and is implied by k 1 l 1 going from minus to plus infinity in fact they are equivalent. So therefore, when we make this correspondence, we notice that we have exactly the same expression. So we have nothing left to do in the proof. We have the same range of summation, we have the same expression and therefore, the left hand side is equal to the right hand side. So let us write it here with this correspondence. The left hand side is equal to the right hand side and hence we have proved what we wanted to do. Now we need to look at the continuous case that requires a little more explanation. The idea is the same, but it requires a little bit of explanation in fact it requires bringing in what is called a transformation of element of integration and here we are talking about a two dimensional integration. So we have to be a little careful let us do that now it is not very difficult, but we need to complete it and there we go. So for the continuous case, let us write down the left hand side and the right hand side again, left hand side x 1 convolved with x 2 and then convolved with x 3 and right hand side x 1 convolved with the result of convolution of x 2 and x 3 and let us evaluate the left hand side at a particular continuous value t and to do that we will first begin by evaluating x 1 convolved with x 2 at the point t very simply it is x 1 lambda multiplied by x 2 t minus lambda d lambda going from minus to plus n and then we have x 1 convolved with x 2 and then convolved with x 3 evaluated at t being x 1 convolved with x 2 evaluated alpha multiplied by x 3 at t minus alpha integrated with respect to alpha from minus to plus n and now we need to substitute for this from the previous expression x 1 convolved with x 2 at t, let us do that. So that is essentially, so now we have a double integral x 1 lambda multiplied by x 2 alpha minus lambda multiplied by x 3 t minus alpha integrated with respect to alpha and lambda so double integral each of alpha and lambda independently go from minus to plus n and let us similarly write down the right hand side. The right hand side requires first an evaluation of x 2 convolved with x 3 at the point t which is very easy let us make it lambda 1 here just to distinguish between the previous lambda in the left hand side and then x 1 convolved with x 2 convolved with x 3 first evaluated t is x 1 alpha 1 x 2 convolved with x 3 evaluated t minus alpha 1 integrated with respect to alpha 1 from minus to plus infinity and that can be rewritten as a double integral again. So there we go we have a double integral again and we need to make a correspondence again between the integrals. Let us identify the correspondence we have them all on the same page we can identify the correspondence. So it is very clear that lambda needs to correspond with alpha 1 I will show them in different colors. So I have alpha minus lambda corresponding to lambda 1. So let me write that down lambda 1 and alpha minus lambda and lambda and alpha 1. So let us put this down in a matrix form the transformation that we require we have alpha 1 is equal to lambda lambda 1 is equal to alpha minus lambda let us write that down the form of a matrix. Now when we want to transform the element of integration d alpha 1 d lambda 1 and express it in terms of d lambda d alpha what we need to do is to consider the determinant of this matrix we need to consider the determinant of this. Now that determinant is very easy that determinant is in fact equal to 1 in magnitude that is what we are interested in we are really interested in the magnitude of the determinant and therefore what we have is that d alpha 1 d lambda 1 by what is called the Jacobian property you know we involve the Jacobian which is the absolute value of this determinant. So either Jacobian principle d alpha 1 d lambda 1 is equal to 1 times d lambda d alpha and of course when alpha 1 and lambda 1 independently go from minus to plus infinity that also corresponds to alpha and lambda independently going from minus to plus infinity and now if we go back to the previous 2 expressions that we had let us see them. Once we make these correspondence the red correspondence and the green correspondence what happens to this the what happens to the argument of x 3 let us put that down in blue. So when we make this transformation if alpha 1 is to be replaced by lambda and lambda 1 by alpha minus lambda then t minus alpha 1 minus lambda 1 would become t minus lambda minus alpha minus lambda which is essentially t minus alpha and that agrees with this and therefore the 2 integrants correspond exactly and we have just seen that because of the Jacobian principle the elements of integration correspond exactly the limits of integration correspond exactly and therefore left hand side is equal to the right hand side and we have proved the associativity of convolution. Now what does this mean convolution is commutative convolution is associative both for the discrete and the continuous case it is a very powerful idea in fact we have used it before this is not the first time we are hearing this but I wanted to go through a formal proof to round things off you know and now what we will do is to say a little more about the typical kind of systems and certain consequences of commutativity and associativity in the next session. So let us meet again in the next session where we shall say a few more things about linear shift invariant systems in general both in discrete and in continuous time. Thank you.