 this argument to convolution. So suppose we take the convolution of two sequences, X1n without the loss of generality has nonzero samples only from only between 0 and n1 minus 1 and X2n has nonzero samples only between 0 and n2 minus 1. In other words, the so-called length of sequence X1n is capital N1 and the so-called length of sequence X2n is capital N2. The question is what happens when you convolve these? What length would it have? Let us see. It is very simple. When we convolve, you see you would have 0 to n1 minus 1 for X1 and X2 is moved around on this. So X2 has n2 samples which move and the movement begins with this sample here. You see the output starts only when this sample reaches here and the output ends when this sample reaches here. In other words, the output would be nonzero only in the range 0 to n2 plus n1 minus 2. You see for this to reach here and to move all the way when this reaches here, you have essentially gone over n1 minus 1 plus n2 minus 1 samples. That is n1 plus n2 minus 2. The movement has been between 0 and n1 plus n2 minus 2. In general, I leave this to you as an exercise. Suppose X1 occupies the range. Now when I say occupies, what I mean is that the nonzero samples are only in this range. Occupies the range capital N1 to capital N2. That is the nonzero samples are only in this range. Similarly, X2n occupies n going from n3 to n4. It should be noted that when we say occupies or when we are saying the nonzero samples are between these numbers, we are not saying that all those samples need to be nonzero. We are saying that if there are any nonzero samples, they are confined to between n1 and n2 for X1 and between n3 and n4 for X2. Some of the samples in between could also be 0. When occupies n1 to n2, X2 occupies n3 to n4, then X1 convolved with X2. That is X1 convolved with X2 occupies n1 plus n3 to n2 plus n4. This is I leave it to you as an exercise to prove. It is simple. Just write down the convolution expression and you would see it fairly in a straightforward way or you could do it graphically. In fact, I would encourage you to use both approaches, the algebraic approach and the graphical approach as well and reaffirm the same result. Therefore, in a convolution of two sequences with n1 and n2 samples respectively, we have n1 plus n2 minus 1 samples. The convolution occupies length X1 plus length X2 minus 1 samples. And this gives us a hint as to how we should deal with convolution in the frequency domain if you want to sample. You see, it is quite adequate if X1n has a length of n1. It is adequate to sample the frequency domain for X1 with only n1 samples. And similarly, if we wish to look at the frequency domain or if we wish to avoid time domain eliciting when only X2 is involved, it is sufficient to take capital N2 samples. But if we convolve X1 and X2 and if we wish to multiply, therefore the discrete time Fourier transform as you would, convolution can be obtained by multiplying the discrete time Fourier transforms and then taking an inverse discrete time Fourier transforms. So, if you wish to use that route, then it is not adequate to sample with only capital N1 samples or capital N2 samples. You need to sample with capital N1 plus capital N2 minus 1 samples on all of them. That is very, very important. If we wish to, let us make a note of that. If we wish to carry out convolution by sampling the DTFT, we must take little n1 or other lengths X1 plus length X2 minus 1 samples in the DTFT. And this is true for all the 3, the DTFT of X1, the DTFT of X2 and the DTFT of X1 convolved with X2. You need to work with this many samples. To take an example, suppose we had X1 of length 3 and similarly for X2, also length 3. Then X1 convolved with X2 would be of length 3 plus 3 minus 1, which is 5. We need to sample X1 and 2 omega with a spacing of 2 pi by 5, not 2 pi by 3. You would need to take 5 samples of X1 omega and X2 omega and multiply the 5 samples, sample by sample and then do whatever you wish to. But if you really take only 3 samples, then what would happen? Suppose we violate this and take only 3 samples of X1 and X2. What would happen? That is, you try and use only 3 samples. You see, how would you convolve? You would convolve 2 sequences by multiplying the discrete time Fourier transforms. See, convolution X1 convolved with X2 means multiplication of the DTFTs. So if you have only 3 samples of this and if you wish to get back X1 convolved with X2, what you are actually going to get, let us call X1 convolved with X2 as Y. What you are going to get is not Y shifted by every multiple of 5 samples but Y shifted by every multiple of 3 samples. So we are going to get summation R going from minus to plus infinity Y in plus 3 R and let us write down this convolution to understand it better. You see, you have X1 0, X1 1, X1 2, the 3 samples of X1 and you would have 3 samples of X2 which move around this, the platform and the train. So the output would be X1 0, X2 0 here at 1. So you see Y of 0 is X1 0, X2 0. Y of 1, you can keep writing it. It is X1 of 1, X2 of 0 plus X1 of 0, X2 of 1. Let us write down all the 5 samples. Y2 is X1 of 2, X2 of 0 plus X1 of 1, X2 of 1 plus X1 0, X2 2 and you can continue. Y3 is X1 of 3 and X1 of 3 is 0 so that does not exist at all. So you see, you have only 2 samples now. You can see what is happening here. When you have 3 then this has reached a 0 sample. So X2 of 1 overlaps with X1 of 1 and X2 of 2 with X1 of 1. So X1 2, X2 1 plus X1 1, X2 2 and finally Y of 4 is simply X1 2 times X2 2. Is that correct? Now carry out this time domain, Elias. What is going to happen? Y n plus 3 times R summed over R will look like this. At 0 you would have usually had just Y0 and then starting from 0, Y0, Y1, Y2, Y3 and Y4 but you are going to shift this by every multiple of 3 and add them. So Y0 is also going to come here. Y0, Y1 and so on. And this is also going to go backwards. So you see it is going to go backwards by 3. So 4 would also come to 1 and as expected this is going to be periodic with period 3. You see the sequence obtained by taking the original sequence, shifting it by every multiple of 3 and adding these is going to be periodic with period 3. That is very easy to show. Let us put n plus 3 here and it gives you back the same sequence. So it is expected that this is going to be periodic with period 3. So we need to take only the principal period of 3 here. Let us call this Y till day 0, Y till day 1 and Y till day 2. And now we have an expression for Y till day 0, Y till day 1 and Y till day 2. Let us write down that expression. Y till day 0 is essentially Y0 plus Y3. That becomes, if you look at the, we derive the expressions for Y0 and Y3. We have an expression for Y0 here and we have an expression for Y3 there. So that is X1 0, X2 0 plus X1 2, X2 2, X2 2, X2 2, X2 2, X2 2, X2 2, X2 2. Y till day 1 is similarly Y1 plus Y4 and that is very easily seen to be X1 0. Let us fix, you know, let us do one thing. Let us write in the same order for X1. So X1 0, X2 1 plus X1 2, X2 2, X1 1 and of course you have, of course we very easily see that Y till day 2 is the same as Y2. Y till day 2 is equal to Y2. And that is of course we have divided before. So we write in the same order as I said. X1 0, X2 2 plus X1 1, X2 2 plus X1 1 plus X1 or other, you know, we could, if you want the same order then we will write X1 2 and then write X2 0 here. So X1 1, X2 1. So you have taken care to write the same order. X1 0, X1 2, X1 1, X1 0, X1 2, X1 1. Here also X1 0, X1 2, X1 1. And of course you notice that it is different samples that are associated with these every time. To understand what is happening better, let us write down the samples not on a straight line but on the surface of a circle. So let us fix the outer circle with the samples of X1. So we have X1 0, go and count the clockwise order. X1 1 and X1 2 and we take an inner circle and put on it the samples of X2. So let us put the samples the way they associate for Y till day 0. So X1 0 associates with, so you know we have taken X1 0, X1 2, X1 1. That is essentially the C-Samp is going in this order. X1 0, X1 2, X1 1. So X1 0 associates with X2 0. So let us put X2 0 here. X1 2 associates with X2 of 1. So let us put X2 of 1 here. X1 1 associates with X2 of 2. So let us put X2 of 2 here. Now when we come to 1 till day 1 these are fixed. So this is the, we call, let us call it the rotor. The inner one we call the rotor and the outer we call the stator. Taking a queue from the terminology for machines. So for 1 till day 1, X1 0 now gets associated with X2 1. There is a movement by one step. You see X1 0 gets associated with X2 of 1. So X1 1 is going to get associated with X2 of 0 as is expected here. And of course X1 2, so the other, you see, so you have, you see X1 0, you move this here, isn't it? For 1 till day 1, X1 0 is got associated with X2 of 1. So this is rotated one step here. So X2 0 gets associated with X1 1, X2 2 gets associated with X1 2. When we come to 1 till day 2, X1 0 is associated with X2 of 2. So this is rotated 2 steps. This is rotated here. This is now rotated. So it is rotated by 2 steps and the associations are after rotation by 2 steps. So it is as if we were convolving not on a straight line. It is as if we had a train and passengers in and out of the train not on a straight platform but on a circular platform. And therefore what we have got here is what we call circular convolution. Call the circular convolution because it is as if you were doing convolution with the sequences put on the surface of a circle not on a straight line. And of course as expected circular convolution is bound to be periodic. I leave it here as an exercise to work out what happens in case you took 4 samples and not 5. So m equal to 4, 5, 6 samples worked for all the 3 cases. Of course of course you took 4, 5 and 6 samples respectively of the sequence X1 and X2 and then worked in the frequency domain. We are assuming that we have a way to go back after sampling but we need to complete that one little step. How do we go back? After we have sampled the frequency axis how do we go back to the original time domain expression? Now we can of course do that only if we have no time domain aliasing. Or even if you have time domain aliasing whatever we do is going to give us the aliased version of the sequence. Either way we must have a way of going back after sampling. Now to do that of course we can again you know exploit or invoke the idea of vectors. So what we are saying is we must now put some terminology. The m samples omega, omega evaluated at 2 pi by n times k, k going from 0 to n minus 1 of the DTFT. These m samples are called the discrete Fourier transform for DFT of the sequence Xn. Now of course we should be more precise. We should call it the n point discrete Fourier transform because we could have taken less for more samples too. And here we are assuming the sequence X of n has utmost n samples. So if it does not of course you could possibly take a discrete Fourier transform but then there would be time domain aliasing or you may be taking more samples than required either way. If you take more samples than required there is no problem. So even if a sequence has only n non-zero samples you could be very well taking more than n samples in the frequency domain. And that you deal as you can see you need to do it when you are trying to convolve two sequences. When you convolve two sequences you need to take more than the number of samples in either of the sequences. It is not unusual to do that. If you have a discrete Fourier transform you need to have an inverse transform. So the inverse discrete Fourier transform of, now we shall use X square bracket k to denote X omega sample that omega equal to 2 power by n times k. The inverse discrete Fourier transform X of k is essentially, you see now here it is very interesting. You have capital X of k to n sample. So it is as if you had capital N dimensions along which you are trying to represent the original sequence. So what have you really done? You have gone from potentially n samples to n samples in a different domain. And each of these samples in the frequency domain corresponds to a vector, a vector created by a rotating complex number, rotating with these frequencies here, 2 by n times k. So in the next lecture we shall see how we would reconstruct the original sequence or at least try or potentially reconstruct the original sequence by using this idea of vectors. Not only that we also see how to do this. You see all this is useful if it is going to make our computation efficient. And therefore after having established that we can reconstruct we also need to see if we are going to get a computational advantage by discretization which we shall also see in the next lecture. Thank you.