 A warm welcome to the 7th session of the third module in the core signals and systems. We have been looking at all the sinusoids that give us the same samples at the same points. And now in this session we are going to put down some guidelines to bring them together what happens if you add them. But before we add them let us first write down the general expression. So, we have this original wave form here the original sinusoid and we have sampled this sinusoid by replacing t equal to n T s and T s is T naught by 4. We are asking all those sinusoids which have the same samples at the same points we wrote down the expression the last time. Let us write down the generalized expression and let us sum them. So, general such sinusoid A naught cos 2 pi into k by T s we have written this before k over all the integers positive integers and the minus goes with the minus and the plus goes with the plus and there is one thing which I am sure many of you must have noticed. I had forgotten this T, I had omitted to write this T in the expression the last time. I hope you noticed it I had left that out intentionally that some of you notice it and point it out and look forward to it is being corrected. Sometimes I intentionally do this I leave a few things uncorrected at that times that you are alert you notice them and you critically review what is being written and here is once that that T was missing in some of the expressions in the last session and now of course I am making it clear. So, there you are anyway that was just a little game as little prank that I played but anyway coming back. So, now you have all these positive multiples of 1 by T s and you add and subtract and therefore, we can now ask what happens when we add what is A naught plus summation over all k what is this let us see. So, if you take k equal to 1 to infinity we do not know if it will converge. So, let us take k equal to 1 to n, n is a positive integer greater than 0. So, we have 2 terms for every element of the summation for every k now you know this is interesting. So, this is of the form A naught cos and you have some term and for some part you take the negative of that term and for the other part you take the positive of that term. So, if you look at this expression it is of the form let us call this part some capital say B, you have the same capital B here and the other part is essentially a capital C and here you have a B minus C and here you have a B plus. In fact, you can write down B and C explicitly let us do that B is essentially and C is essentially and therefore, you essentially have any one term for any one given k A naught cos B minus C plus A naught cos B plus C and this is very easy to sum that is 2 A naught cos B cos C easy to sum. So, let us write down in that form this is B and this is C and as you notice this is also C here the C is independent of k. Now, in fact, you can add this you will notice that you know you can in fact take out this common the C part you can cos C part you can take out common and we will do that. So, this is the sum we need to evaluate now, what is this sum? How do we evaluate? Well, this can be evaluated by a geometric progression. So, we will first split this into two parts and this as half e raised to the power j 2 pi by T s times k t plus e raised to the power minus j 2 pi by T s times k t and of course, you know if you go back to the expression that we wrote previously we should now not forget there is a 2 here, is not it? There is a 2 here the factor of 2 which we had forgotten and we will put it back there now that the half gets cancelled. So, if I bring the 2 here the half would be removed. So, the summation in brackets now becomes 1 plus summation k going from 1 to n e raised to the power j 2 pi by T s times k t plus e raised to the power minus j 2 pi by T s times k t. Now, this is a very simple geometric progression. The only problem is of course, it has a common ratio of magnitude 1. So, you have to be a little careful, but let us take any one of these terms. Now, this can be written as and now you can sum it as a geometric progression. So, let us do that. So, the first term multiplied by 1 minus e raised to the power the common ratio raised to the power n divided by 1 minus the common ratio. Now, we can simplify this. We can take e raised to the power 2 pi by T s times t into n by 2 common and the numerator that leaves us with 2 j sin and the same argument. The same trick can be applied in the denominator extracting 2 j sin 2 pi by T s into t by 2. Now, this is where I am going to leave some work for you to do. The first piece of work is sketch this function of t. Which function are we talking about? This function that we saw, this one. So, sketch this. This is the exercise. Now, you know, you must remember that of course, 2 j would get cancelled here and you can look at this as a sin by sin form. You have dealt with that form before in module 2. So, you know how to handle it. You can also combine all of these terms to make it simple. That should make it relatively easy for you to deal with. So, that is my first job for you, sketch this function of t. The second is what happens as capital N grows? So, the question is what happens? Now, there are 2 things to investigate. What happens at t equal to all multiples of T s? And what happens when t is not a multiple of T s? Now, repeat for the minus term. This is where we are going to be. I am going to leave this as a little job for you to do. I have taken you some distance in the calculation and I want you to carry out the calculation and get a feel. What happens the sampling? Essentially what happens the sampling instance when all these sinusoids come together? What happens at other points, other than the sampling instance? Well, work out something and we will meet in the next session and talk about another perspective on the same discussion. Thank you.