 A warm welcome to the sixth session in the third module of the course signals and systems and in this session we shall continue to discuss the issue that we raised the last time. What are all those sine waves which have the same samples at the same points? I had begun to illustrate with an example, I shall continue that example in this session. I had also given you an exercise at the end of the last session, I am hoping that many of you must have attempted that exercise and probably come up with an answer to, but I am first going to help you by solving it once again and then making some further derivations and some further conclusions from what we did last time. Now, let us put the problem before ourselves again, what we said is you have this sinusoid, so this is the reference here and this is the time axis, by reference I mean where the amplitude is 0, let us sketch a sinusoid and of course, let us assume the t equal to 0 point is here at pi by 4. So, we start taking samples here and we take them with an angular spacing of pi by 2 on this sinusoid. So, we have taken one sample here and the next one here, of course it follows the next one would be here and the next one here and this can continue. So, this was the sampling pattern that we talked about the last time. Now, let us write an expression for this sinusoid. So, let us assume the amplitude is a naught and of course, this interval the period of the original sinusoid is taken to be t naught, the sampling interval is taken to be t s. So, clearly t s is equal to t naught by 4 and the expression for the sinusoid is clearly a naught cos 2 pi by t naught times t plus pi by 4. This is the frequency here, the angular frequency and this is what is called the starting phase, this is the amplitude and we now write down what happens when we miss cycles or miss parts of cycles. So, let us draw the situation, let us expand as I said just one sampling interval. So, I again draw the reference and draw a very big pi by 4 to minus pi by 4 kind of region on this and this is of course, if that is t equal to 0 then this should be t equal to t naught by 4 or t s. Now, as I said you know here this would go on till it reaches and this will also continue till it reaches the maximum, but the point is that you can get the same samples at the same points in two different in fact, in an infinity of different ways, but in two different ways if we wish to just about or actually lose just one cycle. So, let me once again put those before you and I am revising this because it is a very important idea with a loss of essentially one cycle, we have two possibilities. In one possibility we lose the whole cycle and then return to this on the downward edge as it should be which I am now showing in red. So, I go all the way I lose the whole cycle and then I reach here on the downward edge. So, how much of angle have we lost in one sampling interval here? Let us calculate it forgive my drawing it is only approximate, but the idea is if I were to draw a parallel here like this and mark this point then one is losing an angle of complete 2 pi here 2 pi angle lost and then of course, there is a pi by 2 here. So, for the red possibility you are talking about 2 pi plus pi by 2 angle lost in a sampling interval which is 5 pi by 2 angle in the sampling interval TS. So, let us write an expression for this sinusoid clearly the phase varies 5 times and of course, it comes to the same angle pi by 4 at t equal to 0. What is the expression? It has the same amplitude it comes the same phase at t equal to 0 and of course, it changes phase 5 times as compared to the original sinusoid. So, clearly the expression for this sinusoid should be a naught same amplitude cos 2 pi into 5 by t naught plus pi by 4. Now, let us write the expression for the other possibility I had left this to you as an exercise the last time and I am hoping that you must have thought over it, but now I am helping you to do it because it is important enough for us to understand this concept very thoroughly, but I wanted you also to reflect upon it that is very important in a course like this until you reflect and participate with me in working things out you will not grasp them fully. So, it is very important that you work out exercises and then come prepared or at least with some thinking, but anyway let us now complete the exercise. So, here we are we again draw reference mark the t equal to 0 point here and the t equal to ts point here or t 0 by 4 point and then we recognize that there was a sinusoid which took an angle of pi by 4 here and it reached an angle of minus pi by 4 here and it continued downwards here and continued upwards there and now the possibility is that we are losing not quite the whole cycle, but a part of it. So, let us draw that possibility now clear. So, let us mark this clearly here this is where you are this is the situation. So, you would reach this point on the upward edge here reach on the wrong edge so to speak what angle have we covered in this portion we have covered an angle of pi by 2 clear and in this portion we have covered an angle of pi. So, total angle covered is 3 pi by 2 that means you have 3 times the frequency the amplitude is the same. And now what about the angle you see you are now reaching this at the wrong point so to speak you know. Now, how will that reaching at the wrong point be expressed it will be expressed by reversing the angle. So, what is the expression for this sinusoid it is a naught cos 2 pi into 3 by t naught t minus pi by minus pi by 4 this is important. So, you know this minus becomes important because at t equal to 0 there is no problem it is the same cos pi by 4 and cos minus pi by 4 are the same. So, you come to the same point here, but then at this point you get the right value by using minus pi by 4. So, when t is equal now what is t here t is equal to t s at this point t is equal to t s. So, t is in fact t naught pi 4 or equal to t s. So, you have a situation here where you have the same sample reached and you have a different frequency that reaches it with a different phase this is important. So, in both cases you are almost losing a cycle in one case you lose it and arrive with the correct phase in the other case you almost lose it and arrive with the wrong phase. Now, let us write down just these three sine waves for our reference the original one a naught cos 2 pi by t naught times t plus pi by 4 losing one cycle or almost one cycle. Now, you know here we can now write this in terms of t s. So, instead of writing it like this let us write it with a t s there. So, notice something interesting here these two signs agree. So, when I subtract the original frequency from the sampling frequency I also reverse the phase a starting phase and when I add the original frequency to the sampling frequency I add the original starting phase. Now, it is not difficult to generalize to losing more than one cycle or almost losing more than one cycle. In general all the other sinusoids which have the same samples the same points are as follows let us write them down these are the expressions. So, we have these two signs a green and so also these two signs and you essentially have key cycles. Now, it is interesting that all these sinusoids have the same samples at the same points and it is also interesting that all these different cases lose different numbers of cycles. So, naturally the question is what will happen if you put these sine waves together that is an interesting question to answer. Suppose we brought all these sine waves together with the same amplitudes of course, as we have done of at the points of sampling of course, there would be a constructive interference what would happen at other points that is something that we need to reflect upon and we are going to do it in the following session in two different ways. First, I am going to give you an exercise to do to actually add these sinusoids term by term and I am going to show you how you can go about that, but later I am going to approach the problem the other way. Suppose you sort of preempted what might happen how could you prove it? So, let us meet again in the next session the seventh where we shall take up this issue in some greater depth. Thank you.