 Welcome to the 25th session of the second module in the course signals and systems and here we carry our journey further to see can inverse for the Fourier transform. Let us see where we are just to get our bearings right. What we have seen is that if I try to calculate this integral minus infinity to plus infinity h omega e raised to power j omega t d omega multiplied by a constant kappa 0 essentially taking vector intuition that you multiply components by unit vectors along those components and integrate over all the components. This can be rewritten as summation minus to plus infinity kappa 0 and we have another integral inside and just reviewing this because this is a little involved there is no harm in repeating this a little bit and right here it is a recap you need to do a recap and then we rearrange this we said you could keep h t 1 outside and then take the omega integral inside and kappa 0 being a constant we could keep that outside and then look at the integral e raised to power j omega t minus t 1 d omega and this is then integrated with respect to t 1 and we were trying to evaluate this integral and let us look at it once again this was essentially kappa 0 minus to plus infinity e raised to power g j omega t minus t 1 d omega and this could be written again as twice kappa 0 times t 1 you know we said you must take this as a limit so we said we could think of this as kappa 0 limit and this time let me try and improve my notation. So, I will say limit as omega 1 tends to infinity I will improve my notation minus omega 1 to plus omega 1 because you know after all these are values of angular frequencies so although I had written t 1 earlier it is better I use a more suggestive value here and essentially this was related to a sin x by x kind of pattern in fact we know what that sin x by x form is. So, the sin x by x form was as follows you see the kappa 0 remained where it was then you had sin omega 1 into t minus t 1 divided by t minus t 1 can keep omega 1 here and here as well so now we had an x 0 that look like this. So, this is x 0 what you have here as the argument is x 0 so this is of the form to kappa 0 omega 1 omega 1 is a constant as you know sin x 0 by x 0 we can sketch this that is not difficult to do that would have an appearance like this from what we derived in the previous session the only change is the height would now be to kappa 0 omega 1 instead of 1 and you know x 0 here x 0 is omega 1 into t minus t 1 so it is a question of matching and now you see the point is when omega 1 tends to infinity here what is happening really we essentially have oscillations and these oscillations are dying down smaller and smaller look at them. So, essentially we are looking at omega 1 tending to infinity so you can see that this quantity goes to 0 so it is interesting what we find is that you know as long as you have taken a large enough value of these limits of the integral you seem to be going towards 0 it is also function of t minus t 1 you know that is another interesting thing here so what exactly is happening here so if you look at the integral here the integral is troublesome troublesome because the integrand has a magnitude of 1 so you know you have a magnitude 1 integrand being integrated potentially over all the real axis so actually the integral is divergent that is the problem it is not really a convergent integral so where is this going as a function of t minus t 1 then think of it like that it is a function of t minus t 1 and the integral is on omega now let us look at the integral as we evaluated so let us look at this so here as omega 1 tends to infinity this height also diverges where is this null that is also important the null occurs at omega 1 into t minus t 1 is equal to pi that means t minus t 1 is equal to pi by omega 1 as you increase omega 1 what is happening to the sin x by x function for a given t minus t 1 let us look at this expression so here x 0 is always fixed but then what is happening to the area under the main loop that is something that we should reason this is diverging this is not a function in the true sense that in the regular sense the sin x by x function when omega 1 sends to infinity that is the height in the middle is going to infinity so it is not a function real there what is it then so it is suggesting to us movement towards something that we had encountered in the first module namely the impulse essentially this is going to a situation where it captures non-zero area in 0 widths that is the central idea so you see the interesting thing here is that even as the height goes to infinity the width starts reducing towards 0 and finally an asymptotic situation is reached where essentially a certain amount of area is encapsulated all around 0 so if you think of this as a function of t minus t 1 see now we have to change our focus we have to think of this as a function of t minus t 1 not x 0 anymore so let us look at it as a function of t minus t 1 we now think of this as a function of t minus t 1 not x 0 anymore and of course everything is just proportional after all x 0 is essentially omega 1 into t minus t 1 so it is just a scaling so you know on t minus t 1 scale or on the t minus t 1 axis of course this point continues to be 0 but now the null comes at the point a pi by omega 1 on both sides so minus pi by omega 1 on this side and of course these the spacing between the nulls is also the same spacing also pi by omega 1 and the height here is of course 2 kappa 0 times omega 1 so now let us visualize that you could start thinking of this part essentially this part approximate this by a triangle approximate asymptotically by a triangle now if you are asymptotically approximate by a triangle you can find the area of that triangle and you can do the same thing for the other side lobes these side lobes here so you know this side lobe and then this side lobe and then this side lobe and similarly all these side lobes here all these are smaller triangles here and essentially what we have is that the height multiplied by the width which is approximately 2 kappa 0 omega 1 multiplied by 2 pi divided by omega 1 starts approaching a constant for the main lobe that is interesting that is the important part here so the main lobe starts having constant area and that is also true of those smaller lobes on the side so you can visualize a situation where as you go towards infinity you then have a constant area no matter how large you make omega 1 and all this area now gets concentrated around 0 what is this it is an impulse at 0 in fact now you have to worry about what that area is that is why you have the flexibility of a choice of kappa 0 the choice of kappa 0 can be such that that area can be made unity in fact it turns out that if we choose kappa 0 equal to 1 by 2 pi we get this becoming a unit impulse delta t minus t 1 and then we have a very beautiful situation we have integral minus to plus infinity h t 1 times this integral inside no kappa 0 into minus to plus infinity e raise the power j omega t minus t 1 d omega all this business and then d t 1 this whole thing has been replaced by delta t minus t 1 with a choice of kappa 0 equal to 1 by 2 pi and therefore you would get minus to plus infinity h t 1 into delta t minus t 1 d t 1 which is simply h of t so there we are what we have done here is to give a brief explanation of how this orthogonality has worked when you have gone from a finite period to the infinite omega axis you know you need it to worry about what happens when these sinusoids last forever they still remain orthogonal so the orthogonality is in the sense here of an impulse you know it is a more generalized concept of orthogonality the dot product here that you see or that integral which you encounter as a part of the orthogonality issue you know it is not very clear here exactly how this orthogonality has come but it has come as a consequence of the reconstruction formula we will see more about it later but right now in the reconstruction formula you required movement towards an impulse and that is what has happened here if you take a rotating complex exponential all over the time axis well this is a slightly difficult part of the Fourier transform I have explained certain things informally we will see more in the coming sessions thank you