 So, welcome to the 26th session on signals and systems. In this session, we are now going to begin with an example of discrete convolution. So, let us take two sequences and convolve them. So, you saw that in the relationship between input and output that you had a discrete system which was linear and shift invariant and you knew it is impulse unit impulse response. Now, you know we do not have to keep saying unit impulse response every time, we could just say impulse response. The input is xn, the output is yn and yn is known to be summation on k, k going from minus to plus infinity, x of k, h of n minus k. Now, we also noted that this was an operation between two sequences x and h and we will take an example. Let xn be 4 at 0 and then minus 2, 1 and 3 respectively at minus 1, 1 and 2. You remember this notation, this shorthand notation. Similarly, for variety, let us make hn of length 3 that means 3 non-zero samples and let it take the value 5 at 0 plus 1 at n equal to minus 1 and minus 2 at n equal to plus 1 and it is convolve. Convolve is the verb form, convolve x with h. In other words, generate the output y when x is the input to the system and h is the impulse response. So, now, you know, we need to interpret the expression yn in terms of what you are doing to the sequence. So, let us do that. We have y of n is summation k going from minus to plus infinity, xk hn minus k and now we have two indices. You see now for a moment, let us think of k as the index on which I am going to move these sequences. So, you have both x of k and h of n minus k plotted as a function of k. So, here you have k equal to 0, k equal to 1, 2, 3, 4 and so on and you have minus 1, minus 2, minus 3 and so on. So, at k equal to 0, you would put of course x of 0. I am trying to plot x of k first and here you would put x of 1, x of 2 and so on, x of minus 1 and so on. The trick is about h of n minus k for a fixed n. Remember, I am going to fix n now. Obviously, h of 0 would appear at k equal to n. So, you know, it is important when you are trying to write down h of n minus k, it is important to locate the 0 point. That is easy for us to do and the 0 point will happen where n minus k is 0 or k equal to n and then you look at the expression, h of n minus k. So, when I make k equal, you know, suppose I wanted to make this equal to 1. At n of course, when I put k equal to n, I would get 0. When I put k equal to n minus 1, I would get n of n minus n plus 1. So, I would get h of 1. So, what I am saying is, put k equal to n minus 1, it gives n minus n plus 1 which is 1. So, h of 1 appears at n minus 1. That is interesting. So, you have a situation where at k equal to n, you have h of 0 and at n minus 1, you have h of 1 and at n minus 2, you have h of 2 and therefore, at n plus 1, you have h of minus 1. So, what have we done in all the following? At k equal to n, I am going to put h of 0. At k equal to n minus 1, I am going to put h of 1, h of 2 at n minus 2, h of minus 1 at n plus 1 and so on. That means use k equal to n as a fulcrum and put all the samples after that namely 1, 2 in reverse order before that respectively at n minus 1, n minus 2 and so on. And put all the samples before h 0 namely h of minus 1, h of minus 2 and so on after that in reverse order at respectively n plus 1, n plus 2 and so on. So, as you see, what is h of n minus k really? It is the sequence h of k moved first to n and then flipped around k equal to n. Flipping means when you moved it, you would have otherwise kept h of 1, h of 2 after k equal to n, but now you are going to flip them and make them before in reverse order. And you would have kept h of minus 1, h of minus 2, h of minus 3 before k equal to n. Now you are going to put them after k equal to n in reverse order. There is a flipping, there is a mirroring, there is a movement and there is a mirroring. That is what h of n minus k is. Now of course, you can also show that algebraically and I will do that in a minute. So, you could first start with h of k. For a fixed n, h of n plus k means move h k back by n, replace k by minus k means flip or mirror. So, k equal to 0 would of course go to minus n here and on flipping, it would go to plus n. And once you flipped, of course the 0 sample has now gone to plus n and everything else has been flipped around that point. That is what we saw. This is an algebraic explanation. Now let me calculate the output for one point and I would like you to complete the exercise by calculating the output for every other point. So, let me calculate in the example that we had. So, we call the example that we had. We had this example and I am going to calculate the output at y 2, y 2. So, let us calculate y 2 here. Now y 2 would of course be equal to summation on all k, x k h of n minus k with n equal to 2. That is summation k going from minus to plus infinity x k h of 2 minus k. Let me first draw these two sequences. x k, x k looks like this. There is no real change there. And h of 2 minus k looks like this. Of course I would bring the 0 point at 2 and what comes after 0 will come before now and what comes before 0 will go after. So, this is h of 2 minus k. Let me circle them separately. So, h of 2 minus k is this. And what do I need to do? I need to multiply them point by point and add. So, what would I get? How many nonzero products are they going to be? Let us identify them. There is just this product and this product. All other products are 0. This product is minus 2 and this product is 15. That gives you 13. So, y of 2 is equal to 13. Simple. Now I have shown you how to calculate y of 2. Please calculate y at other points and answer the following question for me. How many nonzero points you have in y? Can you make a general inference? If I know the length, that is the number of nonzero points in x and I know the number of nonzero points in h, can I come to a conclusion about how many nonzero points I expect in y? That is a difficult question. But at least for the specific instance of x and h, the exercise for all of you is calculate the entire sequence y, identify how many nonzero points it has and the challenges generalize. We will talk more about this in the next session. But I have shown you how you convolve. Now let me put down the exercise that I had given to you in the previous session a little more explicitly. That is in fact exercises. 1, find y n at all other n. 2, prove that convolution is commutative. That means x convolved with h. Now this is a symbol I am going to use for convolution again. x convolved with h is equal to h convolved with x. Prove that convolution is associated, meaning x convolved with the result of convolving two sequences h1 and h2 is the same as x convolved with h1 first and then convolved with h. I expect you to complete these exercises before we come to the next session where we will say a little more about convolution and the length up to which convolution lasts and the implications of commutativity and associated. Thank you.