 Now let us look at a few more properties of the discrete time Fourier transform. One important property is what happens when we multiply that is in some sense dual to convolution. Let me explain the meaning of the word dual. So in other words if I have x1 and 2 with DTFT is capital X1 and 2. What is the DTFT of x1n x2n that is the question that we ask. Now I will spend a minute in explaining the idea of dual here. We are kind of you know having a foregone conclusion we are trying to use our intuition to come to a conclusion before we actually arrive there. There is one very common principle that is useful in many context in the Fourier transform and that is that very often the roles of time and frequency can be reversed. Now in continuous time and in analog frequency this is exactly true. So one can show that one can more or less exactly reverse the roles of time and hertz frequency in properties. However in discrete sequences and the discrete time Fourier transform this is true in a slightly more extended sense. It is not true obviously. For example what we saw was that when we multiply when we convolve two sequences we are multiplying their discrete time Fourier transform. Now duality would tell us that you could reverse the roles of time and frequency. That means that when you multiply two sequences you expect some kind of convolution to take place in the frequency domain. But it is not obvious what kind of convolution that we will need to do with a little bit of algebra. We broadly understand that we expect some kind of convolution that is what duality says. But we need to put down exactly what kind of convolution. To arrive at that answer let us actually consider a product. So here we are. So consider the DTFT. Now of course we have assumed that both x1 and x2 have their own DTFTs and then this converges. We are assuming otherwise it has no meaning to discuss. All that we do is cleverly replace one of them by their inverse DTFT. So we write down x2n for example. You could do it the other way too but we will write down x2n as now here we cannot write omega because you already have an omega there. So you must use another variable of integration here. Let us use the variable lambda. Is that right? So we simply replace one of the sequences by their essentially write the sequence as an inverse DTFT of its own DTFT. Now we substitute that. So here we are. That becomes summation n going from minus to plus infinity x1n by 2 pi integral minus pi to pi and you see we realize that we could as well take the e raise to the power minus j omega inside here. It does not do any harm at all. It is a constant anyway with respect to this. It is a constant with respect to lambda so I can push it inside and now I can interchange the summation and the integration and I can combine these two terms and there I am. This becomes 1 by 2 pi integral from minus pi to pi summation on all n x1n e raise to the power minus j omega minus lambda times n multiplied by x2 lambda d lambda. So here we have something very interesting. What we have inside the curly bracket here is essentially the discrete time Fourier transform of x1n but evaluated at omega minus lambda instead of at lambda or at omega. You have just evaluated that discrete time Fourier transform at a different value. Is that correct? So there we are. We are saying this is 1 by 2 pi integral from minus pi to pi capital X1 but evaluated at omega minus lambda and x2 lambda d lambda and recognize that this is very similar to what you expect a continuous time convolution to be. So it looks like a continuous variable convolution. There is only one little catch. A continuous variable convolution should have run all over the independent variable axis. So here lambda is a continuous variable. So it should have taken you all over the lambda axis if you were to convolve the continuous functions capital X1 and capital X2. But you are restricting the convolution to one period. You have restricted the convolution to the period minus pi to plus pi. In fact, we do not really have to insist upon the interval minus pi to plus pi even in the inverse discrete time Fourier transform. In the inverse discrete time Fourier transform we can take any contiguous interval of 2 pi. So if you are really very fond of it you might take the interval from 0 to 2 pi instead of from minus pi to pi. That because from pi to 2 pi you have just the same thing that you have from minus pi to pi. If you are even more insistent on doing something unusual you could start from pi and go up to 3 pi whatever you please. Any contiguous interval of 2 pi is fine and that is true here as well. So one must in general say in this expression that you need to calculate this integral over any interval any contiguous interval of 2 pi it could be from 0 to 2 pi it could be from minus pi by 4 to 3 pi by 4 if you like you know or I mean you know whatever you prefer. So my or minus pi by 2 to 3 pi by 2 right it should be actually pi plus pi 3 pi by 4. So it could be from minus pi by 2 to 3 pi by 2 it could be from 0 to 2 pi or it could be from minus 2 pi to 0 whatever you please any contiguous interval of 2 pi is fine right. Anyway this is what is called a periodic convolution it is a periodic convolution of two periodic functions capital X1 and capital X2 remember both these DTFTs capital X1 and capital X2 are periodic functions. Now if you did actually write down the expression of convolution as you have done here but you try to evaluate this integral going from minus to plus infinity of course that integral would diverge because both of the functions are periodic this would be a sum of the integral over every period and if any of them is non-zero that sum would obviously diverge. So it does not make any sense when you have two periodic functions to calculate the convolution by integrating over all the independent variable acts it only makes sense to calculate over one period and that is exactly what we are doing here remember that in this expression both X2 of lambda and X1 of omega minus lambda for any fixed omega are both they are both periodic of course X2 lambda is periodic with period 2 pi X1 omega minus lambda is also periodic with period 2 pi and how do you get X1 omega minus lambda from X1 lambda let us just spend a minute in doing that. So just to take an example suppose you happen to have this from minus pi to pi for X1 lambda I mean let me assume for the moment that X1 lambda has 0 phase so we will say that the angle of X1 lambda is equal to 0 for all lambda and we have shown the magnitude here let us assume that this is the kind of X1 lambda we are dealing with just for simplicity how would X1 omega minus lambda look essentially it would look with this occurring at omega of course always remember that you have periodic continuation here there is periodic continuation there and there is also periodic continuation here also whatever you see here let us mark it as script A is going to be seen as its own reflected version here let us call it a prime and what you see here let us call it script B is going to be seen here script B prime is also a reflection you see this can you can come to this conclusion in two steps when you replace lambda by omega plus lambda instead of omega minus lambda when you replace it by omega plus lambda you are going backward by omega and then when you replace lambda by minus lambda you are making a reflection so what was at 0 would have gone to minus omega when you replace lambda by lambda plus omega and then when you replace lambda by omega minus lambda you are switching the lambda sign of lambda and therefore what is that minus omega now comes to plus omega what is after minus omega goes before plus omega and what is before minus omega goes after plus omega and that is how we come to the conclusion that whatever was after this would now appear before this in reflected form whatever is before this would appear after this again in reflected form this is the relation between X1 lambda and X1 omega minus lambda. Now let us take an example to illustrate this idea let us take the very simple discrete time Fourier transform which is 1 between minus pi by 4 and plus pi by 4 and 0 else and we will assume that the discrete time Fourier transform is 0 phase so in fact let us take this to be both X1 and X2 of omega of course X1 and 2 omega plus 2 pi is equal to X1 2 omega for all omega and we will assume that angle of X1 2 omega is equal to 0 so essentially 0 phase and the magnitude is 1 between minus pi by 4 and plus pi by 4 0 outside let us find out the inverse discrete time Fourier transform of this all right now of course how we find out the inverse discrete time Fourier transform we multiply this pi raised to power j omega n and integrate over omega from minus pi to pi simple. So we have X1 or for that matter 2n is integral from minus pi to pi X1 respectively 2 omega raised to power j omega n d omega divided by 2 pi and this of course boils down to 1 by 2 pi integral from minus pi by 4 to pi by 4 1 e raised to power j omega n d omega this is a very easy integral to evaluate in fact it takes us only a minute to evaluate it this simply becomes now please note I have written this do we need to put a condition here yes indeed what is the condition the condition is that n must not be equal to 0 provided n not 0 now when n is 0 what does it become when n is equal to 0 this becomes 1 by 2 pi the integral is simply 1 d omega so essentially pi by 4 minus minus pi by 4 which is pi by 2 is that correct so therefore this is valid only when n is not 0 so there we are so X1 or 2n is 1 by 2 pi e raised to power j pi by 4n minus e raised to power minus j pi by 4n by jn for n not equal to 0 and it is very easy to see that this becomes sin pi by 4n divided by pi n for n not equal to 0 and c equal to pi by 2 divided by 2 pi for n equal to 0 now we ask what is the discrete time Fourier transform of X1n into X2n what do we expect it to be we need to convolve the discrete time Fourier transform of any one of them with itself so we take the lambda axis we put here of course we you know let us first just draw one of them so this is just one of them and we can visualize the other so let us visualize the other and kind of expanding it a little bit let me draw them on the same on the same scale so to speak now what do we need to do we need to multiply these 2 and to integrate over 1 period and we can choose that period to be between minus pi and plus pi how much is this interval this interval is pi by 2 so let us assume that this tip has reached you see now you must see take the energy of the train and the platform right so you have this the only catch here is now you know instead of the train having people standing at discrete locations people stand all over the train that is the continuous time situation right so here we are here this is these are the people on the platform these are the people in the train omega moves right now suppose this point has reached here what is the situation you know of course this is going to be repeated at every multiple of 2 pi but luckily when this point reaches here we the trouble would have come from the next such the next such square is not it how far is the next such square 2 pi away from here that is very far from so it is even if even if this point is here the next of these squares is not going to clash with this is that right so what we are sure is that one time only one square clashes with one square here that I leave it to you to verify you will never have a situation where 2 such squares clashed with this one square here okay I leave it to you to verify that that follows from the fact that you have only a pi by 2 interval of spread anyway so what would happen as omega moves from a point where omega plus pi by 4 is equal to minus pi by 4 you see when omega when this point is here omega plus pi by 4 is equal to minus pi by 4 in other words omega is equal to minus pi by 2 and you know at that point this you know so you can visualize this comes here and it moves as omega moves in this way so you have more and more of this square overlapping with the square here or this rectangle and as omega moves along this the area increases linearly the area is obviously equal to this hyped into the base which overlaps and that base increases linearly with time so right from the point where this is here up to the point where this has reached here there is a there is a linear increase right afterwards as this point you see so I mean well well no slight correction right from the point where this has reached here up to the point where this so let us go back to this discussion here we begin from the point where this edge has reached here when this edge reaches here up to the point where this edge reaches here there is an increase of area and the area goes to a maximum when this whole rectangle overlaps with this rectangle here afterwards this edge begins to move away and this edge starts to approach this one when there is a linear decrease what is the maximum area the maximum area is when these two rectangles overlap entirely and that area is of course pi by 2 multiplied by 1 so we have the following shape for the convolution so the DTFT of x1 n times x2 n would look like this it would start when omega plus pi by 4 is equal to minus pi by 4 or omega is equal to minus pi by 2 I could end when omega minus pi by 4 is equal to plus pi by 4 or omega is equal to plus pi by 2 at 0 it would take the maximum that is when omega plus pi by 4 is equal to pi by 4 there would be a linear increase here and a linear drop there and this height reached would be pi by 2 this is how the DTFT would look now you know this so called dual result that we have here is very important later when we talk about the effect of windowing on sequences or when we try to design finite impulse response filters with windows we shall gain a lot of insight into what happens when we truncate an impulse response by using this idea of multiplication of sequences so it is not without application or without reason that we are discussing this property anyway so much so for that property but this property at the moment gives us something equally valuable and interesting let us write it down yes there is a question that is right that is a very good question so the question is here we did not run into any trouble or we you know the whole convolution became easier because two rectangles did not overlap at once you know each of those DTFTs is periodic so what we said is I left it to you as an exercise to show that when one of the rectangles was interacting with the basic rectangle between minus pi and pi no other rectangle interfered because the other rectangle was too far away what would happen if this were not the case well if this were not the case you have to account for rectangles that come at once carefully so in fact that is a very good question and let me leave it to you as an exercise to do the following exercise let the two DTFTs x12 omega take the following form let them be 1 between minus 3 pi by 4 and plus 3 pi by 4 and of course angle is 0 0 phase and of course you know that x1 and 2 have the DTFT capital X1 and 2 as before the exercise is in the DTFT of x1 n x2 n in fact I will also leave it to you as an exercise to show that x1 or 2n comes out to be sin 3 pi by 4n by pi n for n not equal to 0 and 3 pi by 4 by pi for n equal to 0 anyway here when we do this exercise we will have to worry about more than one rectangle overlapping at once you have to carefully account for the rectangles that would overlap with the basic rectangle each time you move