 a warm welcome to the 11th session of the third module of signals and systems. We are now proceeding to a point where we know what trouble sampling causes and we have analyzed that trouble by a process which we might want to call divide and conquer. Similarly, we have tried to divide the problem of identifying the clutter caused by sampling by looking at what sampling does to one constituent. What I mean by that is, if you can think of a waveform, a signal as comprised of many sinusoids and what does that mean? When you say you can think of a signal as comprised of many sinusoids, not necessarily discrete ones. I mean the frequencies do not have to be discrete frequencies. They can also be a continuum of frequency. For example, if a waveform is periodic, it has a discrete set of frequencies in it. On the other hand, if a waveform is a periodic and if you can still think of it as a combination of sinusoids, then it would require a continuum of frequencies. We have seen that in module 2. Anyway, what does it mean when you say that a waveform can be thought of as comprised of many constituted sinusoids? It means the waveform has a Fourier transform. Let us write that down as the first point. So, we focus on signals that have a Fourier transform. What does it mean? Let me give you an example of both. You see, take x t is equal to e raised the power minus t u t. u t is the standard unit step. This x t has a Fourier transform. It is also very easy to find the Fourier transform. Let us find it. So, it is easy to do it. You see, all that you have to do is to multiply x t by e raised to the power minus t omega t integrated with respect to t. We have done this before. We have done this in module 2 and that boils down to this expression integrating only from 0 to infinity. So, let us evaluate. This is an easy integral to evaluate. It becomes integral 0 to infinity e raised to the power minus 1 plus j omega times t integrated with respect to t. Then, this is very easy to evaluate. So, this goes to 1 by 1 plus j omega, very easy. It is simple and elegant Fourier transform. Of course, let us sketch this. You know the magnitude. We have done this before. Actually, we have seen such expressions before and just recapitulating some of them. So, the magnitude and the phase, is not it? We can talk about the magnitude and the phase. So, let us plot both of them. The magnitude is 1 by 1 plus omega squared positive square root and that can be sketched like this. Symmetric about 0 and monotonically decaying on either side. Just for completeness, let us also write down the phase. Although we do not directly require it here, we will write down the phase. The phase is minus tan inverse omega. So, that looks of course, as you can see as omega tends to plus infinity, it would go to minus pi by 2 and as omega tends to minus infinity, it goes to plus pi by 2. Essentially, it is a tangent curve. Now, let us go back to the magnitude curve here. Let us notice one thing about this magnitude. The magnitude never goes to 0. This is something that we need to note. You see, I have intentionally taken this example of a Fourier transform. So, here I am building the story slowly and I have an example here of a signal which has a Fourier transform. Unfortunately, that or yy, unfortunately you will see later, but fortunately or unfortunately that Fourier transform never quite goes to 0 for whatever finite omega you consider. Now, let us take an example of a signal which does not have a Fourier transform. We know such examples, but let us recapitulate them. All that we need to do is to make the exponential positive. So, we write e raised to the power of t u t. You remember the signal will look like this. It would grow exponentially without bound starting from 1. Here, the integral minus 2 plus infinity x t e raised to the power minus t omega t dt is divergent. It does not converge. So, this signal has no Fourier transform. Now, let me make it very clear right at this very junk chart. For the moment, for all through this discussion on sampling, we omit such signals from consideration. Let us be clear about. We do not want to make life difficult. We will focus on signals which have a Fourier transform. Now, let us go back the signal which we took a minute ago, where there was a Fourier transform and let us say something about linking it to the idea of constituents. So, take the signal again. x t is e raised to the power minus t u t. We know the Fourier transform which we can write as capital X of omega if you please or j omega whatever. Let us not be too fussy about notation. We can reconstruct x t from capital X omega and you know how to reconstruct. We have done that in module 2. Essentially what you need to do is to take a product of the component multiplied by the vector, integrate over all components. Recall, you know, we had given a geometrical interpretation to this. The component, the vector aggregated over all vectors and a normalization. This was the interpretation. What are we saying? We are saying that x t is essentially a combination of all these complex exponentials e raised to the power j omega t. And of course, now look at x omega and x minus omega. If you focus on this very expression here, x of omega is the complex conjugate of x of minus omega. So, they have the same magnitude and opposite phase. So, notionally we have a combination like this. x omega multiplied by e raised to the power j omega t plus take the corresponding negative value. You get it, right? What I am doing really is just picking so to speak, you know, go back to the previous expression that we wrote. I am just picking here. I am just picking up a particular omega and I am taking that omega and the corresponding negative omega and bringing them together. And I am thinking of the integral as doing this infinitesimally with all such omegas. So, focus on one omega and the corresponding minus omega, then take the next omega and the corresponding minus omega and you can keep doing this all over the positive axis by taking the corresponding points, the negative axis. And all of them come together and bring the signal back. So, I am focusing my attention on one of them. I am thinking of it as a constituent now. Now, let us do that. Essentially, this is like a constituent here. And what is this constituent? If I expand it, the constituent looks like this. And I know the relation between these. Do not die. These are the same. These are the negative of one another. If I add this, I can easily simplify and get mod x omega. Well, e raised to the power j omega times e raised to the power j angle of x omega. So, e raised to the power j omega t times e raised to the power this part, angle x omega. And it is complex conjugate. This, essentially this part would be the complex conjugate of that. So, the quantity and its complex conjugate added together that gives you two times the real part. And how much is the real part of that cosine omega t plus angle x omega? So, this is one constituent, so to speak. And now you have a continuum of such constituents. You require a continuum for omega going from just beyond 0 to infinity. At 0, of course, I agree there is just 1, 0 and minus 0 are the same thing. But at all other non-zero omega, you have these two coming together and forming a constituent sinusoid. You have already written down the angle and you know the magnitude. Now the moment suppose you were, suppose you did not know this function has this Fourier transform. And you sampled it. What would happen? Every such constituent would be sampled. And every constituent would create its own imposters. And I need to be able to distinguish those imposters from the original constituent. Can I do it in this case at all? Can I ever sample adequately so that I can distinguish the imposters from the original constituent? Think about it and we will meet again in the next session to answer the question. Thank you.