 A warm welcome again and Sushrut and I now continue our discussion. So, first I need to complete my answer to Sushrut's question about commutativity and associativity. Let me talk about associativity. Now, let us first understand the meaning of associativity. Associativity as the name suggests means what association we make out of three system. There may need to be at least three systems. So, I have three systems, one after the other. Now I could either associate these two systems together and then let them operate on S1, on the output of S1 or I could associate S1 and S2 together and let them operate on S3 that is let the output of these be fed into S3. This is what associate. Associativity says that it does not make a difference if S1, S2, S3 are all linear shift invariant systems, if all S1, S2, S3 are linear shift invariant then this makes no difference. Now, physically what would it mean? It would essentially mean that you could have two systems of one kind and one system of another. So, you could do very well. See, in fact, you also might have the flexibility. So, I gave you this business of you know, at the abstraction level S1, S2, S3 have the same meaning, but at the implementation level you may have the choice to implement S2 either in one domain, say the mechanical domain or the electrical domain or maybe the hydraulic domain. So, I am giving you an example of physical situations where in the from the same abstraction you could have different realizations. Now, what associativity says is depending on your convenience, you might choose to put two systems into one domain and the third system into another, right. So, you could put S1, S2 into one kind of a domain, a mechanical domain S3 into another or S2 and S3 into one kind of domain S1 into another and that so the way you make these associations, if you keep the order the same has no effect on the input-output relationship. Now, when you put commutativity and associativity together, you get something very interesting and I think I have also indicated this as an exercise for the class. I have said that you can then prove by induction that any reordering, any permutation of the systems leaves the overall relationship unaffected. Now, for two systems, this is trivial, this happens on account of commutativity. For three systems, one can use a combination of commutativity and associativity and prove it. And then for more than three systems, you need to use mathematical induction and in fact, let us encourage all our viewers to do that, use mathematical induction to prove formally. You know, those of you who have found the formal proofs, it will be an interesting thing to do. Use commutativity and associativity together to show that any reordering, any permutation of systems, LSI systems in Cascade leaves the input-output relationship unaffected. Good, Sushrut. So, do you have any other questions? Yes, sir. We talked about the condition for stability and while we got the condition that the impulse response must have a finite absolute sum or absolute integration. Very good. While proving that this condition is necessary, we gave a special input to the system. So, can we logically possibly arrive at that input because, I mean, it works out fine, but sometimes… How did we make that choice? Yeah, we talked, that is a good question. What we said, we talked about a troublesome input, is not it? I think that is the term we use, a trouble-making input to the system, is not it? I think, I do not know whether I talked, I think I talked about that man who knew many languages. Yes, yes. Yes, I talked about him. You know, the person who used to create troubles to the trouble-making input to the system. So, how did we arrive at this trouble-making input? You see, let us look at the input-output relationship, you see. So, we said that the output, let us take, say, discrete time first, that is easier to understand. So, y of n is summation k going from minus to plus infinity x k h n minus k, where h n is the impulse response or vice versa. Now, you know, the way we arrived at that input was to try and force y n to be bounded in spite of trying to bring in an absolute sum here. So, what we were trying to do is, we tried to force an absolute sum here and still hold y as bounded. Now, how can you force an absolute sum here? You need to do two things. You see, you need to somehow take care of that x as a product there. So, you know, if you want to force an absolute sum, you want the output to depend only on the impulse response. So, how will you do it by trying to keep an input which is essentially 1 or 0 in magnitude? You see, you want to keep an input because you do not want the input to create a change in magnitude. So, how do you? You know, what we needed was, we needed to use an input either 1 or 0 in magnitude. So, let us look at that equation again. So, we said y of n is summation over k, all k h k x n minus k and you know, it does not matter what n we choose. So, you might as well choose y of 0 because the same construction can be used for any other n shift invariant anyway. Let us make life simple. Now, here how do we make this input 1 in magnitude? So, to make it 1 in magnitude, essentially you need to look at the and you want to make it 1 in magnitude and you want to remove. So, you want to make a modulus here. So, what you need to do is to remove the angle of this. That is what we did. So, we said put x of minus k equal to h of k divided by mod, rather h conjugate of k divided by mod h k, if h k is not 0 for that particular k and then is equal to 0 else because you cannot divide by mod h k otherwise, you see. So, when you did this, you had an input which was essentially 1 in magnitude, whenever the corresponding impulse response term was not 0 and when the impulse response term was 0, you let it pass as it is. You do not have to worry about it anyway. So, you can just make the input 0 at that. In fact, you could have made the input anything you like at that point. It would not have mattered, you see. But you could as well make the input well, you know, if you made the input anything you like at that point, then you have to ensure the input is bounded. So, instead of make life simple, make it 0, so you are assured that you do not have to worry about those points at all. So, you assure a bounded input by ensuring the bound is 1 and that bounded input is troublesome because it forces that y of 0 to become the absolute sum, that is the idea. The same idea can be carried to the continuous time context. So, I guess, you know, there must be many other questions which all of you have. Sushrut has just sort of touched some of the questions which all of you must be having. I am sure you must send your questions to us and Sushrut will come again in some subsequent discussions. Tell us more about what he thinks should be explained better. Thank you so much.