 Welcome to the 33rd session and here we continue what we were doing in the 32nd session namely to carry out that operation between the two finite length sequences x and h to obtain y. We had obtained only one point in the previous session. But what we were trying to do is to establish a methodology. Let us put down the question once again just to recapitulate and to put ourselves in perspective. Consider with the arrow notation xn given by minus 1, 953 with the arrow at point 0 here and hn equal to 1 minus 1, 2 with the arrow at 0 here. And we had calculated y0 given a summation only over two points. Now we want to calculate a general yn. So we take for example a general n, y of n. Now you see let us write down the indices 2 over a large range. Let us write them down starting with minus 2 and let us write down x of k on top of that because x of k is not going to change. So it is minus 1 here, 9, 5 and 3 there and everywhere else it is 0. Now h of n minus k we need to reason this out. You see h of n minus k will become h0 when n minus k is 0 or k equal to n. When we increment k so for k beyond n we get values before 0 and correspondingly for k before n we get indices after 0. So let us for example put down h of 2 minus you see at k equal to 2 that is here. We put down h of 0 which is 1. For k after 0 we put down indices before. So indices before are all 0 so we get 0s here and for k before n we get indices after 0. So 0 here, 1 there and 2 here. This is how it is. And now we can calculate y of 2 based on this. We essentially need to look at the products of corresponding points. This point with this point similarly this with this and this with this plus 9 into 2 18 plus 5 into minus 1 minus 5 plus 3 into 1 that is 3. So 13 plus 3 which is 16. Now you know you get the general idea. You first put down whatever is at 0 at the point k equal to n. What was after 0 in h is put down sequentially before n in the same order. So of course you know at n you put down what was at 0 in h at n minus 1 you put down what was at 1 at n minus 2 you put down what was at 2 and so on go backwards. Instead of you see as you move forward in the sequence h you put the same values backward in h of n minus k on the k axis. And what was before so of course in this case there was nothing with all 0s before. So you keep putting those 0s after k equal to n. In fact we can see the same thing by a different line of reasoning. Let us now look at that. So for example we could choose to write down h of n minus k as h of first let us write h of n plus k. So h of n plus k is the same as h of k plus m. And remember here k is the index and n is a fixed integer. You are considering a particular value of n. Now h of k plus n is essentially h of k shifted backwards by n. And then we are making a replacement h of k plus n with k replaced by minus k. Now k replaced by minus k is essentially reflection about k equal to 0. So essentially what we are doing is you have this index of k here. Let us write down a few points there. And you make exchange here. We exchange minus 1 and 1. We exchange minus 2 and 2. And similarly minus 3 and 3. And so too minus 4 and 4. It can go on and on and on. So essentially it is a process of exchange. Exchange of corresponding positive and negative points. Now what may happen when you exchange it like this? You have shifted. Remember you have shifted the sequence backward by n steps. So now when you make this exchange, when you make this reflection about the point k equal to 0, effectively the 0 point which had moved to minus n would now go to plus n. And when shifting it backwards you would have kept the points in the same order with respect to 0. But now that order will be inverted. That is what we saw. We saw that the 0 point goes to n. Then the points after 0 go before n in the same sequence and the points before 0 go after n in the same sequence. So we have arrived at the same conclusion by a different line of reasoning. You could use whatever line of reasoning you like. But this is the essential idea. So let us summarize this. Essentially what we are saying is h of n minus k is essentially h 0 brought to n, h 1, h 2 and so on brought to n minus 1, n minus 2 and so on respectively. And similarly, h minus 1, h minus 2 and so on, 2, n plus 1, n plus 2 and so on respectively. Now, you know, let us spend a minute in thinking what we are doing. Now I have taken this for a fixed n. What will happen when I increment n by 1? The whole structure will move forward by 1 step. Increment n by 2, it moves forward by 2 steps. Once you have fixed this for a particular n, when you change the n, you are just going to move as it is. The structure moves as it is. You do not have to redo the structure every time. So instead of, if for example you had constructed h of 2 minus k, you were trying to construct y of 2. So you constructed h of 2 minus k. Now you want to go to h of 3 minus k. All that you do is whatever you were constructing as h of 2 minus k, you just put it forward by 1 step. Or if I wanted h of 1 minus k, I would have simply moved backwards by 1 step. So basically what I am trying to get at is that you could think of this whole h as a train and you could think of the engine lying at 0 and the engine is at position n now. And of course, all the other points are like bogies connected to the engine. The only thing is you have to think of the engine as being in the middle and the bogies on both sides of the engine. You know trains do run like that too. There is nothing wrong in having the engine in the middle of the train and pushing it. Little unconventional, but not impossible. So we have this train with the engine moving according to n and pushing all the bogies with it. That is the analogy which you should think of. And there is a fixed set of points. So therefore, this whole operation that I am talking about, I am now going to give this operation a name and I am going to use what I call the train platform analogy to explain the operation. I shall show you the entire operation for this particular sequence point by point and using what I call the train platform analogy. Let us do that next. In fact, what we shall do is to take this up systematically in a whole session because we want to illustrate this both for discrete and for the continuous independent value. So hold on until the next session where we shall use the train platform analogy to explain this whole idea. But before we go to the next session, let me at least complete one job that I need to. That is to give this whole operation a name. So we will give this operation a name, naming this operation. We will call this operation convolution that is the noun. We have heard this before, but I am just trying to repeat certain ideas. And the verb form is convolved. We will say we convolve two sequences. So we are saying we convolve the input with the impulse response to get the output and the operation that we are performing between the impulse response and the input is convolution. In the next session, we look at convolution in greater depth by using the train platform analogy. Thank you.