 A warm welcome to the last response session in this module 2 and of course also in this first course on signals and systems. So this is the last time I am addressing you live in that sense though of course there are the video sessions there for you to see and understand. But this is a response to some of the very interesting questions that have been raised on the discussion forum in the last few days and in fact I had anticipated those I had sent an email where I had anticipated those questions and I thought I should give some pointers to the answers to the questions that you could think further on the answers. Now let me first come to the discussion forum and I will point out a few questions that have been asked. Akash Kiran 95 has asked these questions I will just read out a few of them. So he has asked how one can assume kappa to be 1 by 2 pi just like that he is talking about the reconstruction of the signal from the Fourier transform and of course my teaching associate Pratik has given a very appropriate answer. Now I thought I should also elaborate upon the answer that Pratik has given and tried to give a different perspective and in fact I will also take some of the other questions that Akash Kiran has raised he has asked how one can integrate over one lobe and then conclude its area is something and you know that was not such a convenient way of finding out that inverse and he has also asked how you could make a continuum as k seems to be discrete. So you know you are moving from the discrete to the continuous here when you take t growing smaller and smaller rather t growing larger and larger and therefore 1 by t growing smaller and smaller. Now I will answer this question first you know how you can take 1 by t or rather 2 pi into k by t as a continuum when k is discrete. So that is straightforward so you see what we said was if you take a Fourier series representation then you are essentially placing frequency components at all multiples of 1 by t and when you think of them as complex exponentials then you have both positive and negative multiples. Now as t tends to infinity that is we are going towards a periodicity 1 by t tends to 0. So these points come closer and closer to one another. Now what you must appreciate here and that is the slight weak link as I call it is that when the points come infinitesimally close in the asymptotic situation then discrete goes to continuous. Intuitively this is not very difficult to understand because we are saying that the points come arbitrarily close and then you can always have this proponent-oponent argument that you can always find the t large enough that if I take two points sufficiently close to one another you can find one of those multiples of 1 by t in between that is the way to see you know in this movement from discrete to continuous you must use a proponent-oponent argument. If the opponent tells you in this tiny interval give me a multiple of 1 by t you can find a t large enough or in fact you can find many multiples of 1 by t if t is large enough. So you there are two things here in an arbitrary small interval not only can you find a multiple of 1 by t if t is large enough but you can find any number of multiples of 1 by t if you make t large enough. So that is the argument that is the way to understand movement from discrete to continuous. So that is the simpler of his questions and I have answered it here and of course these questions I am sure pertain to many people that is why I am answering the questions in this response video. Now coming back to some of the other questions that he has raised in very pertinent ones so you see he says how can you integrate and find an area there or for that matter what is the purpose of that integration I interpret the question like that what is the purpose of that integration and as I said Pratik has answered my teaching associate Pratik has answered very well I would like to elaborate and maybe give a slightly alternate view to the answer so that everybody might find this little easier to understand. So you know let us go back again to that discussion of perpendicularity of two rotating phases you see what we said is if I have two rotating phases e raised to the power j omega 1 t and e raised to the power j omega 2 t and we wanted to find their inner product let us first define them from minus t to plus t and then make t tend to infinity their inner product would be e raised to the power j omega 1 t into e raised to the power minus j omega 2 t d t integrated from minus t to plus t. What is important here is that you have taken the complex conjugate of the second as usual and this can be captured. Now you can evaluate this integral in fact you can evaluate this integral treating omega 1 minus omega 2 as a variable. So you get the inner product to be now you can simplify this. So you notice that this can be written as 2 j sine omega tilde t and I can multiply in the numerator and denominator by t. So I have this now you see what is happening here the expression that you have is t times 2 j sine omega tilde t divided by j omega tilde t and you can also multiply by 2 and divide by 2 and then cut off the 2 j all together. So that leaves you with you know so this is an interesting sketch on the omega tilde axis and think of it as a function of omega tilde let me sketch the function essentially a sine x by x kind of pattern. Now what we wish to see is the integral of this entire function over all omega tilde you see what we are trying to say is of course these points here you know for example this point here this point these are all the nulls and you can mark the nulls this null will come when omega tilde t is pi or omega tilde is pi by t and this would come at 2 pi by t and so on. So as t tends to infinity the nulls come closer and closer. Now what we want to see you see it is very clear that you could integrate the function as it is. So we can integrate this function as it is and let us write down that area let us write down that integral integral to t and here I take the integral from minus to plus infinity. Now I make a very simple transformation I say put omega tilde t equal to another variable lambda. So t d omega tilde is d lambda and therefore d omega tilde is d lambda by t. So I can rewrite this we have a t here and you have a d lambda by t there. So essentially minus to plus infinity to sine lambda by lambda d lambda this is the important point here. So this is the important point to be noted this integral which we see here is independent of t. It has nothing so whether you make t large or small it remains independent and that is what is captured in t going to plus infinity. So as t goes to plus infinity all this area gets concentrated around omega tilde equal to 0 and that is what you mean do not when you say impulse. You see you mean you are putting a non-zero area into a 0 length. Why am I saying all that area gets concentrated around 0 because the main lobes side lobes all become infinitesimally small. So all the area you know as you go away from 0 there is nothing left those side lobes have all disappeared and the main lobes side lobes all have got concentrated into a smaller and smaller and smaller length and asymptotically we can think of all that as being around 0 itself that is why we are saying it is and it goes towards an impulse. Now how much is the area encapsulated by the impulse that is integral would tell us. So if I want a rigorous value for what a rigorous justification for the value of the integral I should try to evaluate this integral sine lambda by lambda. And in fact there is a bit of a chicker and egg problem here you can evaluate this integral by making use of the property the Fourier transform. So now I have given you a hint how you could justify the correct value to buy there and you could proceed from this slightly better argument about this inner product going towards an impulse and try to answer the question that Akash Kiran has raised. Well so much so for this question now a few other things before I conclude my response I am very happy to have seen the excellent participation by many of you in the course. It is soon going to be time for your final examination I would like to take this opportunity to wish all of you the very best for your final examination please do not be frightened of the questions take them very calmly try to answer as many of them as possible and I am sure all of you will do very well. And we all look forward to having you in the next course as well the second course with module 3 and module 4. Meanwhile put in your very best into the final examinations and if there are any other points that you would like discuss please continue to post them on the discussion forum. It has been a great delight to have all of you participate thank you.