 Welcome to the 14th session in the second module in the course signals and systems. In the 13th session, we did a bit of recapitulation, it was a bit of a difficult subject that we were getting into and we needed to get our coordinates correct. What exactly were we doing and where exactly are we restricting ourselves? Anyway, now let us answer the question. We had raised a question that in the previous session. We said now that we have noticed that we have an orthogonal set there, how exactly given such a signal which we wish either to restrict our attention to the interval 0 to t or any interval of t for that matter or we wish to consider a periodic signal, how do we get these components a k cos 2 pi by t k t plus phi, how do we get the a k's and the phi k's? Of course, we know the angular frequencies, but we want to get the amplitudes and the phases which should be put on those angular frequencies to come together and form the signal. Well, the answer is very easy if we again recapitulate that it was an orthogonal set. So, the question is how do we get a k and phi k in the expansion? How do we get a k and phi k in this expansion? Xt is summation k going from 0 to infinity a k cos 2 pi by t k t plus phi k, where Xt is either periodic with period t or we wish to restrict ourselves to Xt for t between 0 and t. Well, let us answer the question by noting that we have an orthogonal set here. We have seen that these two sinusoids, the two sinusoids with k and l as indices are orthogonal. Now, for example, suppose I took three-dimensional space and I had an orthogonal set of vectors in three-dimensional space. I wish to find out the decomposition of an arbitrary three-dimensional vector into the individual components in these three directions. What would I do? I would take dot products, so I can use that strategy. I can take the dot product with a unit vector in each of these directions. So now, here we have an infinity of such directions, the countable infinity, infinity indexed by the integer k, k running from 0 to plus infinity, all the positive or the non-negative integers. So, if I wish to find the component along a k cos 2 pi by t k t plus phi k, I need in some way to find out the dot product between an arbitrary signal which I wish to so decompose into this expansion and a unit vector in that direction. Now, what is a unit vector? You first need to find a unit vector. So, let us do that. So, how do I find a unit vector? I would have to find the magnitude of this quote-unquote vector lying on the interval 0 to t. So, let us find that magnitude. How would I find the magnitude? Well, it is interesting. You can find the magnitude by taking the dot product of that vector with itself. Let us do that. So, magnitude of the vector as we called it, you know, we are thinking of signals as vectors now, restricted to the interval 0 t is the dot product of this vector with itself raised to the power half. You know, when you take the dot product of a vector with itself, this is bound to be positive. This is the square of the magnitude and therefore, we need raised to the power of half that it has to be non, it has to be non-negative. Let us take that dot product. That dot product or inner product is integral multiplied cos 2 pi by t k t plus phi k multiplied by the same thing and integrate it from 0 to t, which is essentially cos squared of 2 pi by t k t plus phi k and this is easy to integrate, you know, this little lemma in trigonometry. Now, here I could split this into two terms in the integral. I could split this into 0 to t half d t, which is very easy to evaluate and the other one is also easy to evaluate. So, here this is very easy to evaluate. This simply gives you half into t and this is clearly 0 because it is the integral over a finite number of cycles over an integer number of cycles. So, overall the magnitude square that we have is t by 2 and therefore, the magnitude would be the square root of this. So, you know, the unit vector that we want is essentially this vector divided by the square root of the square of the magnitude or in other words, the so-called unit vector that we want. So, the so-called unit vector that we want is essentially the original vector divided by the magnitude or the square root of the square of the magnitude, which we will now write down. So, the unit vector is essentially the original vector, I must emphasize unit here cos 2 pi by t times kt plus phi k divided by the square root of the squared magnitude, if you want to call it that. So, now I have a very simple situation. I can take an inner product of the original signal with this unit vector. Now, the only problem is in this unit vector, there is one unknown phi k and you know for different phi ks, the answer is going to be different, the integral is going to be different. So, I need to know what phi k is. Now, how do we resolve this? We resolve this by noticing that actually you do not have here one vector, but you could think of this actually as a vector in two-dimensional space, let me explain what I mean. So, I have here 1 by square root t by 2 cos phi k cos 2 pi by t kt minus sin phi k sin 2 pi by t kt. So, it is interesting, actually I have two vectors not one which come together to form the space span by this unit vector and those two vectors are the cosine and the sine components. These are the two vectors which come together to form that sinusoid again, to form this sinusoid and now I need to check whether these again are perpendicular vectors or not. So, let me do that. Let me take the dot product cos 2 pi by t kt and sin 2 pi by t kt, dot product is very easy to calculate its integral 0 to t cos 2 pi by t kt times sin 2 pi by t kt dt, now all that I need to do is to notice that I can put a half outside and a two inside and this is simply it is simply the integral of a sinusoid over a finite number of or finite number of complete cycles. So, here again we can use the same argument, we are integrating this sinusoid over an integral number of cycles and so the integral has to be 0 and therefore, the cosine the sine components at that frequency are also perpendicular to one another. And now the cos you know if you go back to that expression that we had here, if you look at this expression here, the cos phi k and the sine phi k are essentially constants now, unknown constants if you like and all that I need to do is now to think of the problem of decomposing x t into its component along this particular angular frequency as two problems, decomposing it into the component along cos and the component along sine and I know that cos and sine are orthogonal. So, now my job is extremely easy. I decompose along the cos, I decompose along the sine and then I put them together. How do I decompose along the cos and decompose along the sine? Well, I use the same principle, I find the unit vector in that direction and in fact I know what that unit vector should be. You see in particular you could have put phi k equal to 0 or phi k equal to pi by 2 and you would have had a unit vector for the cos and the sine parts. So, what I am saying is I do not need to work again to find a unit vector. I already have the unit vector with me. Look at the expression here. In this expression putting phi k equal to 0 gives cos, putting phi k equal to pi by 2 gives sine. So, I do not need to find the unit vectors separately. I know what these unit vectors are and let me write that down. So, the two unit vectors are essentially cos 2 pi by t kt divided by the square root of t by 2 and sine 2 pi by t kt divided by square root t by 2. Now, we are all set to find the components in the next session and we will do that. Thank you.