 So, welcome to the 27th session on signals and systems. In the previous session, I had left a couple of exercises for you to do. I hope you have done them and now we are going to understand the implication of those exercises. What does commutativity mean? What does associativity mean? Maybe we have hinted a little bit when we talked about continuous systems, but now we are going to look at both continuous systems and discrete systems together. So, we have said several things about discrete systems now and you can notice that a lot of things are very similar to continuous systems. Now, let us dissolve these differences. Let us take continuous and discrete linear shift invariant systems together. So, let me put that context down very clearly. We are treating LSI systems in the continuous and discrete context together, that is important. And what does commutativity mean? But before we ask what commutativity means, let us ask what associativity means. So, let us consider a sequence xn being given to two systems one after the other, xn or xt. Remember we are treating continuous and discrete systems together. So, this is s. This could be either a discrete system or a continuous variable system. Now, s we will number it, we will call this s1 and we will call the second system s2. And both these systems are linear and shift invariant. The impulse response of s1 is h1 and s2 is h2. So, what do you expect the output to be? y of n or y of t. And in fact, now we would not distinguish continuous and discrete. So, we will simply put down y is equal to x convolved with h1 first and then convolved with h2. Now, here it could be continuous y, continuous x, h1, h2 or discrete y, discrete x, h1 and h2. The relationship is the same and associativity tells us x convolved with h1 convolved with h2 is equal to x convolved with h1 convolved with h2. What is the physical implication of this? We are saying consider a third system s3 whose impulse response is h1 convolved with h2 and give the same input x to it and you will get the same output y. So, what does associativity tell you? Associativity is essentially a statement of equivalence of systems put in series or in cascade. When I put two systems in cascade, you can replace them by one system whose impulse response is equal to the convolution of the first impulse response with the second. And now I can bring in commutativity. What does commutativity tell you? Commutativity says the order of the system is irrelevant. That means if you were to cascade s1 followed by s2, it will be the same as s1 following s2. These are the same. So, associativity tells you that there is an equivalent system and commutativity tells you the equivalent system is unaffected if you reverse the order of the two systems in the cascade or the series connection. In fact, I now have an exercise for you, a slightly tricky one, but I am sure with a little reflection you should be able to do it. I will put down the exercise. Commutativity and associativity together say if I have capital N systems in series connection or in cascade, any permutation of these N systems in cascade gives you the same equivalent system. In other words, together commutativity and associativity tell you that when I connect more than two systems in cascade, it does not matter in which order I connect them provided they are all linear and shift invariant. Now, I must exercise a word of caution. This is true if the systems are linear and shift invariant and in fact let me write that down clearly, not just any systems. These systems are all linear and shift invariant if they are not there is a problem. In fact, let us take an example, suppose you have system 1 where you give x could be a continuous independent variable, I mean a continuous independent variable sigma or a discrete one and essentially the relationship is y is x squared and s2 is essentially the relationship you have x and you have y and y is x plus 1. So, you know if it is continuous time what this would mean is that yt is xt the whole squared and if it is discrete time it would mean let me show discrete time in red y of n is x of n the whole squared similarly here this is discrete then it would be y of n is x of n minus x of n plus 1 and if it is continuous and you are saying y of t is x of t plus 1. Now, in the case essentially you have taken two system both of which do not have memory what do you mean by saying we do not have memory there is a point wise mapping the output at a particular point depends on the input only at that point. Now, let us cascade these two systems you have s1 followed by s2 and the second instance you have s2 followed by s1. Let me show by taking the continuous example as in point. So, suppose I have xt being given here s1 would essentially give me xt the whole squared here and s2 would essentially give me x of t the whole squared plus 1. On the other hand if I have xt here s2 would first give me xt plus 1 and s1 would then give me xt plus 1 the whole squared which is xt the whole squared plus 2 times xt plus 1 and therefore, these are not equal. So, it is very simple to see that when you have systems which are not linear or not shift invariant there is a problem look back at this example here s1 is non-linear in fact so is s2. So, both of these systems are non-linear and as a consequence cascading them does not allow you to change the order of the systems in the cascade. Now, the exercise for you is to come out with other such examples I would like some of you to come out with other such examples and tell us what these examples are examples where one of these properties is violated either linearity or shift invariance and as a consequence when you cascade you do not get the same output in general if you were to change the order or permute the systems in the cascade. So, to emphasize once again when you have a cascade of capital N systems if all of them are linear and shift invariant any permutation of them is acceptable for the overall input output relationship it remains unchanged with any permutation of the N systems in a cascade. But even if one of those systems is not linear or not shift invariant or neither linear nor shift invariant then you have trouble you are not guaranteed that when you change the order of the systems in the cascade the input output relationship remains the same. Now I am going to confuse you even further or maybe in fact remove some of your confusion see remember what we have said here is that when I have systems in cascade and if one of them violates the property of linearity and shift or shift invariance then the overall system cannot be the same always if you change the order. But other instances see we cannot on the other hand we cannot make a blanket statement that the moment you notice systems are either not linear or not shift invariant then changing the order changes the overall behavior that is also not true for example let me take a very simple instance again let us take this instance of y is equal to x plus 1 being the first system s 1 and y equal to x minus 1 being the second system s 2 both these systems are non-linear however these systems are commutative that is interesting. So you could have exceptions what are we saying to be precise we are saying at the moment either linearity or shift invariance is violated you cannot guarantee that changing the order or permuting the order will leave the overall system unaffected you cannot guarantee there could be instances like this where the overall system remains unaffected there may be instances like the previous one where the overall input output behavior is affected. This is the consequence of linearity and shift invariance provided you look at cascades of only LSI systems the moment one of them is not LSI you have to doubt permutation. Now what I am leaving to you as an exercise is to prove formally that if I have a cascade of linear shift invariant systems then a permutation of them leaves the overall system behavior unaffected and I encourage you to do this by using the principle of mathematical induction. This is a slightly of course I have you know it is a little challenging as an exercise I am giving this to you as a slightly difficult problem. Well even if you do not solve the problem entirely you may now accept by taking examples you know if you take 3 systems for example in a cascade you can easily verify that any permutation of them will give you the same overall equivalent system by using commutativity and associative but if you want to prove this in general you would have to use some kind of mathematical induction and that is a slightly difficult exercise. The easier exercise is do it for 3 systems first. Use commutativity and associativity to prove that if I have 3 systems in cascade any permutation of them gives you the same overall equivalent system. If you want to generalize to n systems you have to use a little more of your mathematical induction knowledge and that is the challenging part anyway in the next session we are going to ask the question if I know a system is linear and shift invariant then I know everything about the system given its impulse response. How do I use the impulse response to come to conclusions about the other properties of the system may or may not have namely causality and stability. Thank you.