 Welcome to the fourth session of the third module of Signals and Systems. You will recall that in the previous session we had raised in some sense a dual question namely if I only have the samples and not the a priori information how badly placed am I and we saw that there is a great a terribly high degree of non-uniqueness if we have just the samples and not the a priori information. So, the role of the a priori information is to cancel this non-uniqueness. Now we also said that many signals are amenable to Fourier transformation that means you can think of signals as comprising of sums of sine waves. In fact linear combinations of sine waves taking it to an extreme you take all sine waves whose frequencies are in the continuum in principle all over the real axis or at least over a segment of the real axis and therefore when you sample a signal you are also sampling each of the constituent sine waves. Let me try and illustrate what I am trying to say. I am saying for example suppose we had just 3 sine waves in a signal. So, you are next t which comprises of a 1 cos omega 1 t plus phi 1 plus a 2 cos omega t omega 2 t plus phi 2 plus a 3 cos omega 3 t plus phi 3 and if we happen let us say to sample x t uniformly so what do we mean by uniform sampling? Uniform sampling means sampling with equal intervals take one sample each at uniform intervals. So, for example you could have this situation where you take a sample at 0 a sample at t s t s is the sampling time or sampling interval 2 t s minus t s minus 2 t s and so on so forth. You take a sample at each of these points which means you really have not x of t, but x of n times t s where n takes all integer values we can formally write that like this. You must have got useless notation by now n over all the integers and t s is the sampling time. So, x of n t s is a 1 cos omega 1 times n t s plus phi 1 plus a 2 cos omega 2 n t s plus phi 2 and so on. What I am trying to say is samples of the signal as it were these are samples of the signal and this is an example of the corresponding samples of the constituent sinusoid and there are 3 such constituent sinusoids here. So, all that I am saying is that if I can reconstruct each of those constituent sinusoids I can reconstruct the original signal simple just by adding those constituent sinusoids and in fact the samples of the signal are essentially sums of the corresponding samples of the constituent sinusoids. Let us write that down. So, signal samples equals the sum of corresponding constituent samples. So, you see if I can buy a proper mechanism reconstruct each of these constituent sinusoids then adding these constituent sinusoids together would actually produce the signal back. So, our philosophy in reconstruction is going to be and remember this because it will take us a few sessions to build this whole idea completely. Our philosophy in reconstruction is going to be make an attempt to reconstruct any of these constituent sinusoids provided you can and what do you mean by you can we will understand. Provided you can reconstruct each of these constituent sinusoids and any way these constituent sinusoids had been linearly combined actually each of the constituent sinusoids also has an amplitude and a phase and when you are reconstructing it you are respecting the amplitude and phase. So, if you just add up all these constituent sinusoids you get back to original signal and you do not explicitly have to add that addition is there. The addition is there conceptually in the Fourier transform and inverse Fourier transform you do not have to work to add up you just have to conceptualize the process of dealing with each of these sinusoids. Now remember here I have taken just three constituent sinusoids there is a set of discrete frequencies in general if a signal is not periodic there is a continuum of frequencies. So, when you say constituent sinusoids you have a constituent a continuum constituent yeah requires a little bit of thinking but it is not that difficult. Anyway now let us focus therefore our attention only on one of these sinusoids let me sketch that sinusoids just one of them you know the cosine begins here the zero phase point is here. So, let us assume that I am taking the samples one at where the phase is pi by 4 and then space it again by a pi by 2 gap and again by a pi by you know you have to have uniform sampling you have no choice in that regard this is where the samples come. So, essentially you can see we have taken 4 samples in every cycle of the sinusoids and without any loss of generality let us assume that our zero point is here this is the time axis t and this is t equal to 0 here we are writing down a or a 0 if you like cos 2 pi by t t plus well it has to be pi by 4 because at t equal to 0 the phase is pi by 4 and what is capital t capital t is essentially this let me mark it and this is t s yeah we have got it all right. So, you know what is t you know what is t s you know where the samples are and now we will ask the question first what is the non uniqueness in reconstruction that means if I were to agree that it is a sine wave that I have sampled only one sine wave remember and it is these samples that you record at those specific points what is the extent to which you have created non uniqueness how many other monster sine waves have you created which could have the same sample values at the same points let us in fact draw it on the same graph and let me show you for example that if I wanted to retain these very samples of this sample this sample this sample and so on at these points what are the possible sine wave well one very simple possibility is you could have lost the whole cycle between these two. So, you may have gone down from here gone all through a cycle and then descended to this point you have lost one cycle one whole cycle between samples of course you can see that if you are losing one whole sample one whole cycle between samples here you will also lose one whole cycle between samples from here to here that is not difficult again that would happen between any pair of samples subsequent samples I am talking about immediately adjacent samples now let us consider another possibility and I will show you that other possibility between these two so as not to make the figure too cluttered here I will also show it to you in a different color I will use green now another possibility is you would not quite lost the whole cycle but you are almost losing a cycle in the sense that this is actually coming on an upward edge here so if you were to lose a whole cycle the situation would have been like this all right let me complete that possibility so if I will losing a whole cycle I would continue complete this lose the whole cycle and then rise upwards there this is the possibility where I lose a whole cycle but now let me draw the green possibly the green possibility is where this point of descent see remember this sample also occurs at another point on the cycle and that is this point so this point could be brought here let us show it in green you have caused this to rise and now this is coming on the downward edge you are reaching that point on the downward edge instead of upward I know this is a little confusing let us expand that portion and show you exactly what is happening let me do that so let me once again draw just that segment of the sign I will blow it up hugely I have taken the samples here where I have shown you in red the samples are a phase of pi by 2 apart and now I said there are two possibilities in one possibility I lose a whole cycle which I am now showing in blue so I ascend complete the cycle and then rise to come to this point the other possibilities I do not quite lose one whole cycle but I come to this point on the downward edge instead of the upward the situation is like this which I am showing you in green I do not quite lose a cycle but I am reaching that point on the wrong edge I am reaching on the downward edge instead of the upward now here I have considered just the loss of one cycle well either the loss of one whole cycle or just short of losing one cycle and of course this can happen in every such interval of effectively pi by 2 phase isn't it you can see that you can convince yourself that what I have done between these two adjacent samples can be done between every subsequent pair of samples now what would happen if I lost two cycles the same thing you could lose not one but two cycles and you could arrive at the same point you could lose three cycles and you can arrive at the same point again you could just fall short of losing two cycles by arriving on the wrong edge or you could just fall short of losing three cycles and arrive on the wrong edge so you can lose one cycle and either arrive on the correct edge or the wrong edge lose two cycles and either arrive on the correct edge or wrong edge. So, for every positive integer possibility of losing cycles you have two more possibilities arriving on the correct edge, arriving on the wrong edge. And all these sine waves would have the same samples at the points of sampling. Not as bad as that horrible infinity we saw, but still a countable infinity and infinity akin to the infinity of the integers. Yes, so we need to work hard in the next session. We are going to see how to describe these sine waves mathematically. Thank you.