 Welcome to the 15th session in the second module of the course Signals and Systems. You will recall that in the previous session we had set out to answer the following question. How do we get the components at a particular frequency, sinusoidal frequency which is present so to speak in the periodic signal and let me just put down the two orthogonal vectors that we had identified at that frequency. Let us write them down explicitly. So we said that at the frequency 2 pi by t times k there are two orthogonal vectors so to speak which are 1 by square root t by 2 cosine of 2 pi by t times k small t and 1 by square root of t by 2 sin 2 pi by t kt. Now you know we will use our geometric intuition to understand how we can use these orthogonal vectors to get the components. So suppose you had an arbitrary vector remember you are talking about a two dimensional space now. So we have a two dimensional space at this angular frequency and we want to get the components of this vector. So you know of course that vector could be in a higher dimensional space but I am just showing its projection on the two dimensional space. You know you could visualize it like this. I will draw a three dimensional so called three dimensional vector here. So I have these two components and a third component. Well you know a third dimension and I am just drawing an arbitrary vector like this in three dimension which I am calling v. Now what you need to do is to visualize this vector v as a vector of a higher dimension. So what I am trying to say is that you have this periodic function and you are focusing your attention on one period and that is a vector in a much higher dimensional space out of which two dimensions are present at that particular angular frequency. And what I have shown as v here could be thought of as that higher dimensional vector. So first thing I need to do is to visualize the projection of that vector onto the two dimensional space span by cos 2 pi by t k t and sin 2 pi by t k t. How do I do it? Very simple. I simply drop a perpendicular from vector v. So it comes onto that two dimensional space which is what I am now going to show. So go back to this diagram here. I am just going to drop a perpendicular onto this two dimensional space and I get this vector which I call vp. It is the projection of v onto this 2D space. And now of course we can project this into the two perpendicular components which are the sin 2 pi by t k t 1 by square root t by 2 and cos 2 pi by t k t 1 by square root t by 2 component. You see get this clear. So I am trying to project vp onto this and this. But projecting vp onto this and this is the same as projecting v onto this. Get this clear. You do not have to first drop a perpendicular and then project that projection onto the component. You do not have to do that. However, you could understand the projection of v. You see when you want to find the components of v along these two perpendicular vectors, unit vectors that we have identified, you could do it in two ways. You could first drop a perpendicular and project the vector vp and then project vp onto these two perpendicular components. Or you could just straight away project v itself on the two perpendicular components. It makes no difference. That is because the perpendicular part which you have lost in projecting on the vector vp is anyway not going to come into that projection. You see what I mean is when I take the dot product of v with any of these unit vectors that we have identified, it is the same as taking the dot product of vp with any of those two unit vectors because vp plus let us go back to the diagram. So vp plus this perpendicular component which I am showing here. Let us call it v perpendicular. vp plus this perpendicular component is let us write that down on the next page. v is equal to vp plus v perpendicular and the dot product or the projection of v on these two, projection of v on these two perpendicular vectors which we have identified is the same as the projection of vp on those vector. So for our understanding, we could first project onto vp and then project that onto the two perpendicular vectors. But in computation, we shall just project v straight away, get this right. So I am now going to show diagrammatically how we are going to project by using vp as the vector to be projected. But in computation, we just straight away use v. Let us get down to business. So you have this arbitrary periodic pump. You see now you have let us start. You have vp here. vp, this is a two dimensional space, two dimensional space at this frequency, at this angular frequency. And I have these two perpendicular vectors here which I am going to show in red. What shall I do to find the components of vp? I will take the dot product or the inner product. So I will take inner product of vp. Let us call this vector u1 cap and this vector u2 cap. So I will take the inner product v1 vp times u1 cap and let that be equal to vp1. And similarly, let the inner product of vp with u2 cap be equal to vp2. So therefore, I would now get vp is equal to vp1 u1 cap plus vp2 u2 cap. However, the interesting thing and that is what I am now writing in green is the dot product of vp with u1 cap is the same as the dot product of v with u1 cap and so also for u2. So in other words, when we begin with the function itself original periodic function, let us call it xt with period t. How do I get the particular component ak cos 2 pi by tk t plus phi k? How do I get this? I get it very simply by taking the dot product of xt with 1 by square root t by 2 cos 2 pi by tk t and multiplied by the same unit vector plus very simple. You see, you are taking the dot product with each of the unit vectors, you have the components and then multiply by the corresponding unit vectors and add and that gives you the component at that frequency. That frequency corresponds to a two dimensional space. Now, we are going to make one more simplification. Look at this expression here. Let me identify the constants. You see, this is a constant 1 by square root t by 2, 1 by square root t by 2. Now, you do not really have to have two constants here. You can combine these two constants and bring them outside and therefore, instead of trying to do this, you could either bring this constant into the summation into the inner product or you could take this constant out of the inner product. Either way, it is alright. So, we could write this down as follows. ak cos 2 pi by tk t plus phi k is just very simply. How do you take the dot product? Very simple. Integrate over a period of t. It could be without any loss of generality 0 to t, x t cos divided by t and then multiplied by the same sinusoid. If you like, we could make this t1 that we do not confuse the element of integration. So, now, we have identified any one of those sinusoidal components. And now, of course, we can construct all the sinusoidal components for positive integer k. Just one little detail needs to be completed. What do we do for k equal to 0? Let us see that. So, for k equal to 0, sin 2 pi by t times k t is equal to 0. We have only one vector there. And therefore, a 0 cos 2 pi by t times 0 t plus phi k is essentially a constant. And this is equal to 1 by t integral x t1 dt1, essentially the mean of x t over a period. So, we can get the so-called 0th component by just taking the mean. So, now, we have all the wherewithal to calculate the sinusoidal components of a periodic signal. In the next session, we shall take an example and illustrate how we do this. Thank you.