 A warm welcome to the 19th session in the third module of signals and systems. We have been looking at what happens when you do not quite sample with ideal impulses. And let me quickly summarize the way we have approached the problem, we are going to complete the discussion of what happens in the spectral domain in this session. So we said that if you had this train of non-ideal pulses rather you know pulses that are not quite impulses but which last for an interval capital delta in the whole sampling interval of Ts. So in every interval of Ts you have this pulse, the pulse lasts for an interval of capital delta. And essentially what you are doing is to take a band limited signal or the signal that you want to sample and multiply it by this periodic waveform. We call this periodic waveform P of t so to speak a train of pulses now not impulses. You see the important thing is that the Fourier transform of this has an appearance like this we saw that. Obviously the Fourier transform of P t is again going to have a train of impulses. Any periodic waveform has a train of impulses as its Fourier transform. And the train of impulses that you get in the Fourier transform is obtained by an envelope which we derived the last time. Essentially the Fourier coefficients give you the strength of each of course this is not a uniform train it is not. By a uniform train you would mean a train of impulses all of which have the same strength but here these impulses do not have the same strength. The strength of the impulses relates to the Fourier coefficients here. So the impulse strength so it relates to the Fourier series coefficients. In fact the situation would be like this on the omega capital omega axis the angular frequency axis you would have an impulse lying at every multiple of 2 pi by Ts. The strength of this impulse would be proportional to C0 the strength of this one to C1 and so on. And we also saw the nature of C0, C1, C2 and so on. We saw that Ck has the form gamma 0 sinc. So gamma 0 could be the constant sinc of k times delta by Ts. And of course you could make gamma 0 depend you know you could essentially try and make it independent of delta. I am repeating some of these ideas you know you must fix these ideas in our mind. It is important to review them from various perspectives. So you know it is really samples of an envelope. So I have an envelope like this a sinc envelope and you have samples of this envelope with a spacing of delta by Ts. Now you can visualize the situation as delta by Ts tends to 0. We essentially concentrate around 0. So all the samples start getting crowded here. But when delta by Ts is not 0 we have something interesting. So with delta by Ts not equal to 0 but small enough. How small enough? You know the idea is you want to be able to make use of that main loop. That is the meaning of small enough. You should not have a situation where you take one sample from the main loop and the next sample comes from the side loop. That is not good for you. You want the first few samples to come from the main loop. So small enough so that the first few samples come from the main loop. Let me zoom this to show you what I mean. So C0 is like this, C1 is like this, so is C minus 1 and then C2 and so on. C minus 2 and this. So you know the first few samples come from the main loop. What you observe? The first few samples have decreasing magnitude. Now this is going to be important to us. This decreasing magnitude is going to be important to us. So in fact now let us write down formula. Let us write a Fourier series expansion. That is very easy to do. It is summation k over all the integers. I am sure you must have now got used to this notation k belong to the set of integers. And when we multiply xt by pt, what are we getting? We are getting. Now you see this is a very interesting expression because easily take the Fourier transform of this. We can take the Fourier transform term by term. Why can we do that? We can do that because the Fourier transform is linear. So we can take the transform term by term and then add the terms. Now what is the Fourier transform of one of those terms? That is very easy to see. It is essentially mine. Now of course we assume the Fourier transform exists. We have agreed to do that. Very easy to do this. Very easy to integrate. You can of course combine the exponential terms. So you see suppose the Fourier transform of xt is capital X of omega. What is this expression? This expression is essentially capital X as a function of omega minus 2 pi by T s times k. So it is as if you have taken the Fourier transform and shifted it by 2 pi by T s times k here. You see that is interesting. X omega shifted by 2 pi by T s times k forward and of course remember k can be positive and negative. So it is associated with the coefficient. So you see it is very interesting. Each of these terms actually gives you a shifted version of the Fourier transform. Now you know you will wonder in fact this is exactly what we have been saying all this while. There is nothing new. We have taken the original spectrum and we have shifted it to every multiple of 2 pi by T s and we have added up these shifted versions. That is what we are saying. Let us go back and see. So you see you want to take the Fourier transform term by term. So let us take this expansion here. You found the Fourier transform of this and now the Fourier transform of this can be found by simply multiplying by ck and then you add up all those Fourier transforms taking the summation on all k. So simple. Three step process. We have done this step. Now we multiply that step by ck and then add over all k. Let us do exactly that to obtain the Fourier transform of xt times pT. So therefore the Fourier transform of xt times pT is essentially summation k over all the integers ck capital X omega minus 2 pi by T s times k. What are we saying effectively? We are saying shift capital X of omega by 2 pi by T s times k. First step of course forward for every integer k. Multiply the shifted version by ck. Add up all these shifted versions and there you are. That gives you the Fourier transform of xt multiplied by pT. We have now formally said something which we have been saying by various kinds of illustrations over the last few sessions and in the initial few sessions of this module. We will see more in the next session. We will continue from this point onwards. Thank you.