 A warm welcome to the 12th session in the third module of signals and systems. So, in the last session, we worked with signals that have a Fourier transform and signals that do not have a Fourier transform and we recapitulated some ideas about such signals, both classes. We also agreed to confine our attention to those signals that have a Fourier transform. Of course, we could generalize later, but as far as our discussion on sampling goes for this course, we shall confine ourselves to that class, signals that have a Fourier transform. And amongst those, we looked at the signal e raised of an exponentially decaying signal so to speak, which had a Fourier transform and we understood that essentially that signal could be thought of as comprised of several constituent sinusoids. Let us write that example down from that perspective again. So, we essentially looked at the signal xt equal to e raised to the power minus tut and we recognized that the Fourier transform looked like this and we said, we could think of xt as a combination, essentially of mod x omega times cosine omega t plus angle of x omega. Now, we can fill in the values here, so this would become, these are the values we knew them from the last discussion that we had and what we need to do is to integrate this all over omega. So, limit of a sum over all omega of such quantities that essentially goes towards an integral. In fact, we do not need minus infinity here, we can do with 0 because we have already taken plus and minus, you see now where is the problem? The problem is that no matter what sampling rate you take, focus your attention on this constituent. So, let us write down things again. Go back to the discussion in session 10. There we had seen that if your constituent frequency is 1 by T0 and the sampling frequency is 1 by TS, then the impostors created, of course you know the frequencies created on sampling are as follows. They begin with 1 by T0, then you have 1 by TS minus 1 by T0, 1 by TS plus 1 by T0, 2 times. So, 2 by TS minus 1 by T0, 2 by TS plus 1 by T0 and so on. And of course, the first difficulty comes from here. So, these two can clash first and therefore, we saw there that you needed to have 1 by TS minus 1 by T0 greater, strictly greater than 1 by T0, which meant that the smallest 1 by TS that you could use should be greater than 2 by T0. Now, go back to the signal that we are discussing now. Firstly, when I have a combination of synusoids, can I generalize this constituent rule? So, for that we have to ask a basic question about sampling. Is sampling a linear operation? Let us ask that question first. Well, the answer is yes, of course. What is sampling? The proof is very simple. What is sampling after all? Sampling means multiplying, I mean I am talking about ideal sampling. Sampling ideally means multiplying the signal by uniform train of impulses. How does that train look? There is an impulse at 0, there is an impulse at TS and you can continue this at every multiple of TS. Let us call this uniform train of impulses Pt. So, essentially sampling of Xt means multiplication of Xt by Pt. So, simple. And therefore, if X1t is sampled, of course, we are talking about the same sampling rate, sampled at a rate 1 by TS. It is essentially X1t times Pt. Similarly, X2t sampled at 1 by TS is the product of X2t and Pt. And now take any linear combination, it is very simple. So, any linear combination alpha times X1t plus beta times X2t sampled, of course, this is in the same way, sampled at 1 by TS is essentially the same signal multiplied by Pt which can be expanded. So, in fact, we proved the result here. What have we proved? We proved that by sampling the linear combination, it is equivalent to the same linear combination of the samples. So, therefore, a linear operation, sampling is a linear operation. That is good to see because now we can draw a conclusion about the imposters that get created when you have many constituents. If I add up many constituents and sample them, it is the same as first sampling each of the constituents and then adding up the sampled versions. So, by adding any one of the constituents, I am going to create these imposters 1 by T0, 1 by TS, minus 1 by T0, 1 by TS, plus 1 by T0 and so on. And for every T0, this is going to happen. So, when I put these different T0s together, all these imposters are going to get created. What does linearity assure me? The imposters that are created as a consequence of sampling, a signal with multiple constituents can be analyzed by looking at each constituent individually. Find out what happens because of each constituent and the effect of all these constituents together is just the sum of the effects of each of these individual constituents. Let us put that point down very clearly. So, what we are saying is linearity assures us. So, the effect of sampling a signal with multiple constituents is essentially the linear combination you can even say sum because you are just summing the multiple constituents, sum of the effects. So, for example, if you have a 1 by T0, 1 and a 1 by T0, 2 and a sampling rate, the constituents are these. I am talking about the constituent frequencies and the sampling rate is 1 by TS. Then the imposters as we might call them are as follows, 1 by TS minus 1 by T01, 1 by TS minus 1 by T02, 1 by TS plus 1 by T01 and so on, a combination of all these imposters. So, now we can ask a basic question. What is the one thing that you need? So, that you are in a position to distinguish the imposters from the constituents. You have agreed the signal has a Fourier transform, but given that it has a Fourier transform, is it possible to still keep the imposters distinguished from the constituents for all such signals? Obviously, the answer is no. In fact, we can answer that question very quickly. Only as long as the sampling frequency 1 by TS is strictly greater than 2 times 1 by T0, can we make this distinction? And of course, if your sampling frequency is finite, it means that every constituent in your signal must have a finite constituent frequency. Otherwise, there is no question of being able to distinguish the imposters from the original. That is, the imposters may become smaller than the original in all other cases. So, what are we talking about? We are talking about a signal having an upper limit to its constituent signal frequencies. That is such a very important conclusion. So, what we are saying is, unless a signal has an upper limit on its constituent frequencies, imposters would not be distinguishable. Imposters may become smaller in frequency. So, unless a signal has an upper limit on its constituent frequencies, the imposters may start having a frequency smaller than constituents after a point. That is, 1 by TS minus 1 by T0 may become smaller than 1 by TS. So, we do not want that to happen. Otherwise, you know, the situation is such that we would like to consider as constituents. Now, all frequencies up to a certain value. So, we are going to allow. The only question is, can we allow up to infinite values or do we need to limit? The answer is very simple. Of course, we need to limit. Otherwise, we would need an infinite sampling rate to be able to deal with all these imposters. So, informally, what have we concluded? Essentially, we have said there must be an upper limit on the constituent frequency. And essentially, what we call such a signal in the literature, in signals and systems, is a band limited signal. So, we are saying a signal must be band limited for it to be sampled without these imposters creating trouble. And the second thing is that the sampling must be adequate, whatever that means. Adequate in what sense to take care of imposters. So, we have concluded this session essentially by noting that we need to have two things. We need to have a signal which has an upper limit on the constituent frequency. And we need to have a sampling rate which can take care of all these constituent frequencies. In the next session, we shall formalize this into a very, very fundamental and a very, very important celebrated theorem in signals and systems. Let us meet in the next session to carry out that task. Thank you.