 Welcome to the 24th session of the second module in the course signal and system. We continue with our discussion of the new transform that we have introduced in the previous one namely the Fourier transform. Let us look at the details. So, we said the Fourier transform of the impulse response h t was described by capital H omega given by the integral from minus to plus infinity h t erase the power minus t omega t dt and we thought of this as a component along erase the power j omega t of the impulse response h t and we were asking can we use vector inclusion to reconstruct h t from h omega and we expect essentially that you would have to do a sum replaced by an integral. So, sum over components needs to be replaced by integral over component. Now what are the components here? The components are different angular frequencies. So, you expect that you multiply the component by a so called unit vector. Now how do you get a unit vector from a vector? You need to scale the vector. So, we will allow for a scaling constant. We do not at the moment know what is the notion of a unit vector here because we are talking about a function that lies over all the real axis. So, we do not quite know about unit vectors so far. Anyway component h omega multiplied by unit vector and integrated instead of being added identify each step of this very carefully. Integrated from minus to plus infinity and a constant. A constant could be a function of omega or may not be a function of. In fact, we will find it is not a function of omega but anyway let us write down a constant. Let us call it constant kappa 0. Kappa 0 a constant for making a unit vector out of erase the power j omega t. So, this should give me back h of t. This is what we called inverse Fourier transform. This is how far the intuition took us. Now we have to start doing a little bit of hard work in terms of actually writing down the expression for h omega and then seeing what exactly needs to be done. So, let us substitute here in this expression. We have summation rather integral minus to plus infinity h omega times kappa 0. We will allow for a constant kappa 0 erase the power j omega t d omega and we know that h omega itself can be written and writing that in green now as integral from minus to plus infinity h. Now instead of writing t we will write t 1 because we have to distinguish from the other variable t here. Now when I make that substitution I have a double integral h of t 1 erase the power minus j omega t 1 d t 1 and then kappa 0 erase the power j omega t d omega. We have a double integral now. Now you see in this double integral we will first put the whole integrand together and then we will see if we can do a little bit of rearrangement. Can we exchange the order of the integral and see something in it. The first thing is let us aggregate the integrand. All this is the integrand. This and this together form the integrand and these are the elements of integration. Let us quickly put them all together and we realize that kappa 0 might possibly depend on omega. We do not know. May just depend. Let us see if we can interchange the order. Now you know a word here. If one wants to be very rigorous about dealing with such double infinite integrals. One has to take recourse to certain fundamental principles in real analysis and functional analysis. I am not doing that here because I am not really trying to be that rigorous in dealing with these integrals. I just want to give a functional understanding of the whole argument. So we will assume that h and all the quantities involved satisfy the requirements for interchangeability and we will do the integrand. We will first integrate with respect to omega and then we will integrate outside with respect to t1. And we will see what we need to have for the integral on omega for the reconstruction principle to hold. So now let us look at the expression once again. Now this after interchange would become minus infinity to plus infinity. Now I keep the t1 quantities outside and I take the omega quantity and I can put those quantities together as you see. Now d omega first and then dt. So you see here we said kappa 0 needs to be a function of omega in general. As I said I am trying to give a functional understanding. I am not being too rigorous. We do not have to make kappa 0 a function of omega. We will do that only if required. So let us begin by seeing if we can avoid making kappa 0 a function of omega. With that what is going to happen? So let us observe. Let us look at this particular inner integral here. What we will do here is assume kappa 0 is independent of omega and then evaluate the simplicity. Evaluate but by limits. And what do you mean by by limits? Let us see. So let us start with evaluating minus t to t kappa 0 erase the power j omega t minus t1 d omega. And remember kappa 0 is a constant. So I can pull it out. Now this integral is not at all difficult to evaluate. In fact we can do it right away. This simply becomes kappa 0 divided by j omega. Well you are trying to evaluate as a function of omega. So j into t minus t1. So omega is removed. And you have erased the power j omega t minus t1 from minus t to plus t. Now of course here capital T is a value of omega. So perhaps we could change the symbol if you like. Let us call this minus t1 to plus t1. Since t1 and minus t1 are actually values of frequency, angular frequency. That is not at all difficult to evaluate. So let us substitute that. And this is also very easy to evaluate. This is essentially 2j sin capital T1 t minus t1. Divided by j into t minus t1. And of course I can divide and multiply by 2. And then the 2j goes away. That leaves me with 2 kappa 0 and you can also divide and multiply by capital T1 if you like. So you have sin t1 t minus t1 divided by t1 t minus t1. Now this is an interesting function that we have. We have a function of the form sin x by x. This function is a very frequent visitor in signals and systems. Let us now look at that function in some depth. So let us just focus our attention on the function sin x by x if we have here. You like sin x0 by x0. This one of course is undefined that x0 equal to 0 but you can take a limit. Limit as x0 tends to 0 of sin x0 by x0 can be shown to be 1. There are different ways of proving this. One can use L'Hopital's rule. One could do it by any other way one finds convenience. You could take the derivative of the numerator and denominator and evaluate the derivative divided by derivative at the same point x equal to 0. So if you look at the derivative with respect to x0, it is cos x0 divided by 1 and substituting x0 equal to 0 gives 1 by 1. At other points we are going to have null. So nulls occur at the nulls or zeros of sin x. So if you look at the expression, this is how it could look. Let us plot them separately. The sin looks like this and so on. This is at pi, every multiple of pi. This is sin x0. And the same graph you could draw x0 itself. x0 is like this. A straight line with an angle of 45 degrees or pi by 4. And we divide the black by the green. And you can see first of all that this is going to be an even function because both of the individual numerator and denominator odd. So when you divide an odd function by an odd function, it is going to give you an even function. Moreover, essentially you are dividing by an increasing quantity. So what is going to happen essentially is that the oscillations are going to get reduced. The oscillations are going to get smaller and smaller as you go along. So let us complete the session by drawing sin x by x and coming back to it in the next session. So let us draw sin x by x in total. Sin x by x is going to look like this. It is going to be 1 as you know at the point 0. And then it is going to have oscillations that become smaller and smaller. 1 here and nulls at all multiples of pi. Now we will use this in the next session to draw some conclusions about the quantity that we are trying to evaluate. Thank you.