 In that sense, now we understand how unique this property is for sine waves. You know, when you add two sine waves of the same frequency, you get a sine wave of the same frequency. In fact, now let me give you a justification of that. I will leave the details of the justification to you, but if I have two sine waves of the same frequency, a1 say cos omega 0 t plus phi 1 plus a2. Now, I am even giving them different amplitudes, but the same frequency. I am also giving them different phases. I can always decompose each of these, isn't it? So, I can decompose them. Let me decompose them into their cosine and sine components. So, I have this as a1 cos phi 1 plus a2 cos phi 2 times cos omega 0 t minus a1 sin phi 1 plus a2 sin phi 2 times sin omega 0 t. It is a simple trigonometric exercise to combine this bag again to get a term of the form a3 cos omega 0 t plus phi 3. I leave it to you as an exercise to complete the details. That means, find a3 and phi 3. What are they? It is a simple exercise. I mean, it is a simple exercise in basic algebra, trigonometry, whatever you want to call it. Anyway, this proves this beautiful property of sine waves. So, now, there is one good reason why we like to build our thinking around sine waves. We very often think of, as I said, all reasonable signals as a combination of sine waves. We like to think of them like that. In fact, you may wonder if sine waves are so smooth. Smooth means they have an infinite number of continuous derivatives. In fact, there is a term used in the literature for such functions. They are called analytic. They have an infinite number of continuous derivatives. If sine waves are so smooth, how is it that they come together and form rough functions? And we know that, you know, they do that to a certain extent and they do create trouble at discontinuities. But we will not get into those details at the moment. What we do know and we have been exposed to this idea from our introductory courses on signals or signal theory or our introductory courses on transforms that we can use what is called the Fourier transform to decompose or to regard a continuous signal xt as a possibly infinite sum or linear combination. I will continue this as a possibly infinite sum or linear combination of sine waves of frequency varying in principle from 0 to infinity. I repeat that because it is an important idea. The Fourier transform regards a continuous signal xt as a possibly infinite sum or linear combination of sine waves of frequency ranging all the way from 0 to infinity. Now, of course, if the waveform is inherently periodic, let us for the sake of argument, take the period to be 1 millisecond, then we have associated with that waveform what is called a fundamental frequency of the reciprocal of 1 millisecond namely 1 kilohertz. And in that case, you do not need all these frequencies from 0 to infinity. You need only discrete frequencies. You need the frequencies 0, 1 kilohertz, 2 kilohertz and all multiples of 1 kilohertz. On the other hand, we could take this argument to its limit. So, suppose you have an aperiodic waveform, a waveform where the period is notionally not finite and that holds the key to generalization. So, if you have an aperiodic waveform and if it is quote unquote reasonable, I cannot qualify that at this time. Let us assume and let us be satisfied with the explanation that we will only deal with reasonable signals most of the time. But anyway, if the signal is reasonable and if it is aperiodic, then you could think of it as if it had a period of infinity. And that means its fundamental frequency would tend towards 0. And in contrast to the case where you had a 1 kilohertz fundamental frequency and you needed only frequency which are multiples of that. Now, you need multiples of what is potentially a 0 fundamental, which means you need essentially the continuous frequency axis. So, for aperiodic waveforms, we need to move from a discrete frequency axis to a continuous frequency axis. Anyway, you see it makes it clear why we are in a position at least to deal with all reasonable signals if we can deal with one sine wave. If we can see what happens to one sine wave when we sample, we will be able to come to a conclusion on what happens to any reasonable signal when we sample. So, now let us answer the question. Let me take just one sine wave. So, let that sine wave be A cos omega naught t plus phi or if you like A naught cos omega naught t plus phi naught. And let us sample this sine wave at t equal to all multiples of the period capital T. Let us give this signal a name. Let us call this sine wave x of t and therefore, we are essentially taking x of n t. And we shall again use a slightly compressed notation for this. We will say x of n t is x square bracket n. Let us make some observations. It is very easy to see that if you had x of n t or x square bracket n as we have just called this, which of course, now would become A naught cos omega naught n capital T plus phi naught. You could as well have got this from the expression A naught cos omega naught n t plus phi naught plus 2 pi times n k where n and k are integers. Any integer k here would give you the same value. That is not at all difficult to see. The same time instant or time instant index if you would like to call it that, small n and different integers k. All of them are going to give you exactly the same value for every point small n. This is something rather striking. In fact, what it tells you is that we seem not to be answering the right question. When we sample a sine wave, what we have landed up doing is creating too much of ambiguity or too much of confusion, not loss of information as we thought initially. We are asked the question how do we ascertain that we are not losing information by sampling. Perhaps we have to rephrase the question. By sampling what we have done is to create a lot of confusion. There is too much of information now. These samples could have come from so many different sine waves. There was just one sine wave initially. You took samples of sine waves based at intervals of capital T. Now when you do that, your problem is not that you cannot reconstruct a sine wave. You can reconstruct a sine wave. There are just too many sine waves that you can reconstruct. In fact, now let us put down explicitly what those sine waves are. In fact, let us increase the confusion in a minute. So let me go back to this expression here. I have a0 cos omega0 nT plus phi0 plus 2 i nk. Now you know I can invert the sine. So I know that cos says psi is equal to cos minus psi. So you know I have trouble there. I can introduce a minus sign here and get the same values. So I can put a minus sign in this expression here. Therefore, because of this I can also equate this to a0 cos well plus minus 2 pi nk if you like to call it that minus 2 pi nk minus omega0 nT minus phi0. But you see k took all integer values. So since I am free to choose all integer values there, I do not really have to write minus k. I can also write just plus k there. So I will do that. I will write plus k here. So now I have a very serious problem to deal with. I have a0 cos omega0 nT plus phi0 is equal to a0 cos 2 pi nk plus omega0 nT plus phi0 and it is also equal to a0 cos 2 pi nk minus omega0 nT minus phi0. In other words, if I were to aggregate terms together, collect terms as they say, we are saying essentially that x of n is equal to a0 cos omega0 nT. So I will write it as omega0 divided by 1 by T times n. You will see why a little later. It is also equal to a0 cos, you know. Now I need to take the n term together. So I need to write it like that, you know. I need to write omega0 by 1 by T times n plus I will write 2 pi, you know, I need to write nT here. So I will multiply and divide by T and of course plus phi0 and that is also equal to a0 cos minus omega0. So in fact let us take this term now. From this term I will multiply and divide by T here. So I have 2 pi by T times nT k minus omega0 nT minus phi0 and all of them give me the same value. So essentially if I were to combine, let me combine a few terms. So for example let me combine these 2 terms of the same place. I will combine these 2 terms. Now you know you can take n common from here. So let us combine terms. I get xn is of course the obvious a0 cos omega0 nT plus phi0 which is also equal to a0 cos 2 pi by T times k plus omega0 into nT plus phi0 and that is also equal to a0 cos 2 pi by T into k minus omega0 nT minus phi0. All 3 of them and now we have a beautiful interpretation that we can work with. Let us look at this expression, this series of equations a little more carefully. What are we saying in this series of equations? We are saying that this stream of samples x of n could have as well come from the original sine wave whose frequency is capital omega0. It could have as well come from another sine wave whose frequency is capital omega0 plus 2 pi by T times k and k can be any integer. So k could be 1, 2, 3, minus 1, minus 2 anything that you like. And finally of course it could also be the frequency 2 pi by T minus omega0 and notice one thing when you associate plus omega0 with 2 pi by T into k you have the same phase as the original sine wave. On the other hand when you associate minus omega0 with 2 pi by T times k you have the opposite phase. So omega0 becomes minus omega0 here. You have a change of phase, you have a reversal of phase. It will be much easier for us to appreciate this if we take some numbers. So let us take some numerical values. Let us again go back to our example of T is 1 millisecond and we are saying essentially that 1 by T is 1 kilohertz then. And let us take omega0 to be 2 pi. Now remember omega0 is an angular frequency. So it has to be 2 pi times a hertz frequency. Let us take it to be essentially corresponding to the hertz frequency of 0.1 kilohertz in kilohertz. Now you know just for the sake of convenience we would now work only with the hertz or kilohertz frequencies rather than a mixture of angular frequency and hertz frequency. So let us use the understanding of 0.1 kilohertz for the original sine wave and 1 kilohertz for the sampling frequency or the rate of sampling. So we are saying original sinusoid of 0.1 kilohertz and sampling frequency or the rate of sampling is 1 kilohertz. And now we sit down to analyze what we have just concluded. We have just concluded that if you take the frequency axis now or the kilohertz frequency axis initially you had only 0.1 kilohertz. So here you have 0 frequency. Here you have 0.1 kilohertz. Let us mark 1 kilohertz there. Let us mark 2 kilohertz somewhere here perhaps not quite to scale but anyway representative. What we are saying is that initially you had a 0.1 kilohertz frequency with a phase of phi or phi naught and amplitude a naught. After sampling what you have created is many, many copies of this or in other words you have created so much of confusion that all of a class of sine waves could have generated those samples. Which are those sine waves? Let us mark them on the frequency axis with their amplitudes and phases. So the same samples could have come from a sine wave of frequency 1.1 kilohertz amplitude a naught phase phi naught or it could have come from a sine wave of 1 minus 0.1 that is 0.9 kilohertz with an amplitude of a naught but a phase of minus phi naught please note. It could equally well have come from a sinusoid of frequency 2.1 with an amplitude of a naught and a phase of phi naught or it could have come from a sine wave of frequency 2 minus 0.1 which is 1.9 kilohertz with an amplitude of a naught but a phase again mind you of minus phi naught and this pattern is repeated. So around every multiple of the sampling frequency 3 kilohertz 4 kilohertz 5 kilohertz 100 kilohertz 1000 kilohertz and what have you you have 2 adjacent frequencies. You have the sampling frequency plus 0.1 kilohertz with an amplitude of a naught and a phase of phi naught and you have this other one with a frequency of sampling frequency minus the signal frequency or 0.1 kilohertz with an amplitude of a naught and a phase of minus phi naught. All these sine waves could have created these samples. So this takes it very clear what we said a minute ago the problem is not that we have lost some information in sampling the problem is we have created too much of confusion in sampling. Now there is just too much to deal with there are just too many sine waves that could have given me the same samples. In fact we need to spend a minute or two in understanding graphically what has happened graphically what has happened is the following. You see let us take the original sinusoid and let me expand just one period. Let us assume that we have taken one sample here and the subsequent sample somewhere let us say here. The confusion has arisen because of the following reason. The same samples could have come from another sine wave where you have lost a complete period in between. So you could have lost a whole period and you could be going back. You could have also lost a period and you remember there is a correspondence between this and this point on the sine wave. So now you could also draw another sine wave where this sample appears on the upward edge and this sample also appears on the upward edge. So you could compress it even further. Here of course you are drawing the sample you are putting this sample as it is originally on the downward edge. But you could even complete this and draw it on the upward edge and then of course you could be skipping two whole periods. You could be skipping three whole periods and again whenever you are skipping one period you could either be putting the second sample on the upward edge or the downward edge. That is another way to understand why for each possibility of multiple of the sampling frequency. So like you had 1 kilo hertz, 2 kilo hertz. You know your original sine wave was a 0.1 kilo hertz, the next one is 1 kilo hertz and 1 plus 0.1, 1 minus 0.1. Now what is the ambiguity between plus 0.1 and minus 0.1? It is essentially in a way the ambiguity of phase whether you have plus phi 0 or minus phi 0. That is a mathematical way to describe it. A graphical way to understand it is at the same point could have come from an upward or a downward edge as you see here. I encourage you to take the exercise of drawing a few of these possibilities and let me conclude this lecture today by asking you to do so. Draw a few of these possibilities so that you get a complete feel of the ambiguity that is created by sampling. Find out many different sine waves graphically that can generate the same stream of samples and you know how to do it. Essentially you get one pair by skipping one cycle, you get another pair by skipping 2 cycles, 3 cycles, 4 cycles and you know if you do it a few times for a few cases the idea would be very clear to you. So there we are. We have created so much of ambiguity by the process of sampling. What we now need to do is to resolve that ambiguity and the ambiguity can be resolved if we go back to the original drawing here. You know the ambiguity has come because of these false sine waves that have been created. We need to understand when it is possible for us to remove these false sine waves and retain the original true one and when there could be some trouble in doing so. We shall discuss this in the next lecture. Thank you.