 A warm welcome to the tenth session of the fourth module in signals and systems. We are now slowly getting familiar with the properties of the Laplace transform the z transform. And the most important property I think which we are now reasonably familiar with at least in the continuous independent variable case of the Laplace transform is the convolution property. Let us now state that property formula. So, let us say property 4. It says that convolution in the natural domain leads to multiplication in the Laplace domain. All that we can say is that the region of convergence of the product of the two Laplace transforms includes the intersection of the regions of convergence of the original two Laplace transform. We cannot really say more in general. Let us first establish the same property for the z transform and then we will elaborate little bit more about this business of intersection. So, what is the convolution property for the z transform? It takes of n as usual have the z transform given by capital X of z with the region of convergence r sub x and the same holds for h of n. What is the region of convergence and what is the z transform of the convolution? What would it be by expression? Let us give this convolution a name. Essentially, it would be summation n going from minus to plus infinity x. Well, you know you want to use a different variable. So, you can call it m going from minus to plus infinity x m h n minus m and this is y of n. This needs to be z transformed. So, you need to take summation n going from minus to plus infinity and put a z to the power minus n there and sum over n. Let us expand this now. As usual, we need to make a transformation of variable. So, we keep m as it is, but we put n minus m equal to l. Thereby, n becomes m plus l and for fixed m, if n goes from minus to plus infinity, so does l. So, this breaks down into summation l going from minus to plus infinity summation m going from minus to plus infinity x m h l z to the power minus m plus l and we can break this down into z to the power minus m times z to the power minus l. We can now separate this into a product and once again, it is very clear that if a particular z belongs to the intersection. So, if a particular z belongs to the intersection r x intersection r h, this product definitely converges for that z. So, therefore, if you look carefully what we have here is the z transform capital H of z. What we have here is the z transform capital X of z and we are sure now that we have at least the intersection in the region of convergence. So, we can now state property 4 for the z transform. When two sequences are convolved, their z transforms are multiplied. What about the region of convergence? The region of convergence of the product of the two z transforms is at least the intersection, if not more. Now, here again, we need to see a little more. See, for example, if there is a non-null region of intersection, at least we are assured there is something. The product will have some region of convergence, we are assured of that. Suppose there is a null region of intersection in the original product. That means x z has a region of convergence r sub x, h z has a region of convergence r sub h and r sub x and r sub h have a null intersection. Then what do we do? Of course, the statement is correct that the product would still have a region of convergence at least equal to the intersection, so at least a null set. But that says nothing. Is it really the null set? What do you mean by having a null set as the region of convergence? Let us see. Let us take the example of just 2 raised to the power of n, not 2 raised to the power of n u n. Consider x n is 2 raised to the power of n for all n. Now, we can write this down as 2 raised to the power of n u n plus 2 raised to the power of n u minus n minus 1. So, essentially what we are saying is separate out the points n from minus infinity towards minus 1. So, take all this together and put it here and then take n equal to 0 onwards and put it here. So, you know in principle when you take the sum, if you use linearity, you should be able to write down the individual z transforms. So, I have 1 by 1 minus 2 z inverse which corresponds to 2 to the power of n u n with mod z greater than 2 and here I would have 2 raised to the power of n u minus n minus 1 which would have minus 1 by 1 minus 2 z inverse, but mod z would be less than 2 and if I just blindly add them, I would get a 0. But remember this intersection is also null. So, the problem is if a z transform has a null region of convergence, it means there is no z transform. So, for example, here x of n equal to 2 to the power of n for all n has no z transform in the traditional sense. If I bring in impulses, the story is different, but it has no z transform in the traditional sense. Now, what could happen at times is that this business of expansion can actually take you to a bigger region of convergence. Again, I am not elaborating fully upon this at this stage, but I am just pointing you to that possibility. Although when you multiply x z and h z, they may as such have a null region of convergence, but their multiplication brings you a region of convergence which is not null, the convolution process. And of course, one thing is definitely there. The region of convergence can always expand beyond the intersection. Let me now take a case as an example where the region of convergence expands. So, let us take x of n again to be our traditional 2 raised to the power of n u n. And let me convolve it with h of n given by the finite length sequence 1 minus 2. So, you know how to read this finite length sequence. This means that h of 0 is 1, h of 1 is minus 2 and h of n is 0 except for these 2 cases. Let me use the train platform analogy to convolve. So, you have the platform on which starting from n equal to 0, you have 1, 2, 2 square and so on sitting and you have the train and the train moves. And you can see that the beginning of any output of the convolution will only be here at n equal to 0 before that it is going to be 0. You can see the output is going to be 0 all this while up to here. At this point it will be 1 into 2 minus 2 into 1 which is 0. At this point it will again be 1 into 2 square minus 2 into 2 which is 0 and you can see that you will get a 0 all afterwards. So, the output is simply delta n and therefore, x of z is 1 by 1 minus 2 z inverse with mod z greater than 2. Here h of z is 1 minus 2 z inverse and the only thing you have to exclude is the 0 point mod z greater than 0 otherwise everything else is all right. Whereas, if you consider the output y convolved is x convolved with h then y of z is 1 and the region of convergence is all z it is the impulse can you see that. So, here the region of the intersection was essentially mod z greater than 2, but the output convolution had a region of convergence all over the z plane. The region of convergence expanded beyond the intersection. Now, I leave it to you as an exercise construct a similar example in the continuous independent variable domain. Can you could be a little tricky a bit of a challenge I leave it to you as a challenge. Anyway, we will see more in the next session about convolution and system functions. Thank you.