 A warm welcome to the 8th session of the third module of the course signals and systems. We have done some math in the previous session. In that session, we have taken all these synosoids which had the same samples at the same points and we had brought them together and added them up. And we had written down a process for adding, we had essentially established an expression when they came together and I had left a little job for you to do to evaluate that expression and plot it as a function of t, particularly noting what happens at the points of sampling and other than the points of sampling. And I must warn you, I hope you might have already noticed that at the points of sampling you have to take limits, you have to be careful, you know because the denominator goes to 0. So, there is a bit of cheating I did there because you know the denominator went to 0 and I still wrote down a summation and the correct way to evaluate that expression at the sampling points is either to use the L'Hopital's rule by taking the limits of the numerator and denominator, you know essentially taking the derivatives of the numerator and denominator and then taking a limit or else by evaluating ab initio separately for the sampling instance and for other than the sampling instance. I hope some of you tried that, but anyway instead of carrying that discussion further I thought we look at the problem from a slightly different angle. Now, you may wonder why I am doing it from this angle, but it is not entirely unrelated after all we are sampling. So, suppose we did the following, suppose we actually physically did sampling and what do you mean by doing sampling? I have the sinusoid, I retain the sinusoid for a very narrow period around the point of sampling and then remove the sinusoid for the rest of it, let me sketch that process. So, what I am saying essentially is I have this original sinusoid, you remember I took samples here at an angle of pi by 4, this was 0, this was Ts, 2 Ts, 3 Ts and so on. Now, what do you mean by taking sample? You could visualize it as if you retained a very small part of the sinusoid around that sample and remove the rest. So, the red waveform is what I have and not the black one anymore. How do we express this process of retaining just a teeny-weeny portion around each sample for the sinusoid and removing all the rest, bringing all of the rest to 0? We are essentially multiplying this sinusoid by a train of very narrow pulses. Let us sketch those pulses. So, what we are saying is multiply the original sinusoid by this train of pulses, very narrow pulses, symmetric pulses if you like, you know where I am getting, is it not? This is kind of suggestive about what we did in module 1 somewhere. The only thing is that here are very narrow pulses have a width of delta, but we are not making the height 1 by delta at the moment. So, you know we are not making the impulses in that sense, we are just multiplying by very narrow pulses. Let us be content with that for the moment. So, let us do that. Now, if you are going to multiply by a periodic, this is a periodic train of pulses. How do I express this operation in an alternate sense? The periodic function can be thought of as a combination of sinusoid. Now, this is of course, this periodic function satisfies the Dirichlet condition. So, you can expand it as a Fourier series. In fact, I have made life easier by taking a symmetric train of pulses, symmetric around t equal to 0. So, therefore, it is going to be an even function of t, is not it? So, let us write a Fourier expansion for it. We can make a Fourier expansion. And we can make the Fourier expansion by identifying the fundamental period here. So, this is T s by 2 and minus T s by 2. This is the period T s. The function is periodic with period T s. Let us write down. Now, of course, since it is an even function, clearly this is an even function of t. So, it is going to have a Fourier expansion of the following form. One constant plus summation k going from 1 to infinity, only cosine terms. The fundamental multiplied by k T. And of course, the phase would be 0, you know, because it is even. We are talking about an even function of T. So, k equal to 1 is the fundamental and k greater than 1 are the harmonics. So, this C 0 is essentially the average of the waveform over the interval. How much is that average? It is of course, 1 multiplied the area under the waveform, 1 multiplied by delta divided by T s. So, delta divided by T s. And how do I get C k? Well, that is easy. Just multiply by cos and then integrate. Very easy integral to calculate. Of course, k is greater than equal to 1 here. Let us substitute the limits. So, of course, we can evaluate this difference and not yet simplifying it too much. As you notice, this has two times of the sinusoidal argument. Now, this is interesting. Again, you have all these going away there. Now, you know, here we have to do a little bit of work. We have got so far and we have the Fourier series coefficients. But now what we want to do is to make the pulses narrower and narrower. And then we will also see, if you make the pulses narrower and narrower and keep the height at 1, the pulse is ultimately going to vanish. So, there is going to be nothing left. In fact, you can see that from the expression here. If we leave this as it is, this term will go to 0. There is nothing left of that term because, you know, it just says that there is no pulse at all and there is no, the whole waveform falls flat. It becomes 0. So, this is the Fourier expansion simply becomes 0. There is nothing great. So, you know, here we need to bring in the same notion that we did in the first module. I do not want that waveform to disappear. Even though I want the pulse to become narrow, I want the Fourier series to be non-trivial. How can I do this? The only way is to make the height grow. And how will I make the height grow? By making the height of the pulse inversely proportional, the width of the pulse and what do we have there? Yes, you are right in impulse. So, this is where that impulse needs to come in. We did not, as I said, we did not start with the impulse. But now I need to bring in an impulse if I do not want something very trivial. So, let us do that. Let us accept. Let us be willing to do that. Let us now make, so situation is like this. We made the height equal to 1 by delta. So, I have the situation that I put an impulse at every multiple of nts height 1 by delta and width delta. And of course, if I do that, all that I need to do is to multiply the Fourier series expansion by 1 by delta. So, let us do that. I simply need to multiply this by 1 by delta. And of course, I can divide and multiply by ts to make matters easy. So, I will divide by ts and multiply by ts all in the denominator. This is getting a bit cluttered. Let us rewrite that. So, what I am saying essentially is the kth Fourier series coefficient now becomes 1 by ts times 2 by pi k delta by ts sin k pi delta by ts and now we have some sense now delta tending to 0. In fact, delta by ts tending to 0 makes sense. In fact, we know what happens. You know, sin x by x as x tends to 0. We know what it is. So, in fact, you see, now look at the expression here as this argument here. So, you see now delta by ts tending to 0 makes sense because you have a sin x by x kind of form where x tends to 0. Let us precisely evaluate that sin x by x form. So, let me write this expression down explicitly. I have 1 by ts 2 by k delta pi by ts into sin pi into k delta by ts and now this whole thing can be taken as an x. So, I have 2 by ts times sin x by x with x tending to 0. So, this tends to 2 by ts. Interesting. So, it is independent of k. That is the important thing. The fact that it is independent of k reminds us of something that we did in the previous set, does not it? You had an expression similar to the previous session which we have now got by proceeding the other way and we shall say more about this in the session to come. Thank you.