 It is obvious that we would like to have a uniform structure repeated if possible and that uniform structure could be degree 1 if we can make it so. Now what we are going to do is to conceive of the simplest possible degree 1 structure that we can envisage and then see whether that degree 1 structure can be repeated to get any rational causal system function. We will find that we can do almost that but not quite. There are a few pathological cases where we might not be able to do this but otherwise in most cases we shall indeed be able to decompose it in the way that we have just described and that leads us to what is called the lattice structure. Let me put down right in the beginning the motivation for the lattice structure and attempt to use a combination of degree 1 blocks to realize any causal rational system function. This is not as exciting as it is. Look carefully at the direct form 2 structure. The direct form 2 structure in a way is that you are using degree 1 terms after all. I mean you are using one delay at a time if you think of it that way and if you look at a degree you know if you look at a direct form 2 what you are doing is you are putting one loop for each additional term in the denominator or numerator. Although this is exciting it is not exciting enough. What I have also said in the beginning of the lecture is that in addition to being able to realize it should also exhibit certain system properties. That is where the lattice structure scores over any of the other structures and we will state that very clearly. In addition the lattice structure makes obvious certain system properties. In fact location of poles with respect to unit circle and why poles not only poles of location of zeroes of poles that is what is attractive about the lattice structure in addition to that degree 1 repetition. So in fact before we you know we cannot really at this point understand all of this we are just trying to put down a wish list and then see if the wish list is realized but what we will do right now is first to think of the simplest possible lattice structure that we can have. Now you know let us the smallest degree 1 structure that you can envisage should have at least one multiplier and one delay. Now there again you want to realize both a numerator and a denominator. So the list you can do is to have two ranches so to speak one which will give you a denominator kind of term and the other which will give you a numerator. Now all this is kind of intuitive I mean in a way what I am doing is I know what I finally have and I am trying to arrive there by some kind of intuitive reasoning. So do not worry too much if we do not understand all of this right now after we have gone through the lattice structure in its entirety everything will fall into place. So what I am saying is that minimum degree to the degree 1 structure probably should look something like this you know you need to have a delay somewhere. So let us put a delay let us put a delay here. I am showing two inputs and two outputs I will explain why you need to have a multiplier you need to have some summation and some multiplication let us put that down let us put say k times or minus k does not matter k if you like. Now you know you could have made done with this but as I said later we would want to tweak this structure to give feedback as well. So if you have feedback in one direction there must be some feed forward and some feedback. So let us keep everything feed forward for the moment but we will allow for one of them to get converted to feedback. So let us keep this as k here again as a multiplier and one there. So this is what we call this is the minimum degree 1 structure that we are going to be with and let us this is the you know this is going to be repeated and let us write down the Z transforms of the sequences everywhere. So this is let us assume that this is the nth stage or the nth such degree 1 structure. Therefore the sequence which has come up to here is the so called n minus 1th sequence. Let us call the sequence here E n minus 1 in the Z domain and we will call this E till day n minus 1 in the Z domain and this is of course going to be called E n Z and E n till day Z and since this is the nth stage we should also number these coefficients here we will call this the nth coefficient k n and k n there as well. Let us write down the relationships between E n and E n minus 1 from this drawing clearly E n Z is equal to E n minus 1 Z k n times E n minus 1 till day Z I will just show that to you again E n Z is now see be careful this is E n minus 1 Z the truck brings it and transports it here and puts just E n minus 1 here this truck and this truck together carry E n minus 1 and k n times Z inverse E n minus 1 till day Z times Z inverse so I need a Z inverse there too let us not forget the Z inverse. So Z inverse E n minus 1 till day Z appears here multiplied by k n what about E n till day E n till day takes k n times this so k n E n minus 1 plus Z inverse E n minus 1 till day transported as it is. So we have E n till day Z is equal to Z inverse E n minus 1 till day Z plus k n times E n minus 1 Z playing the very beginning this is the nth stage n can be 1 2 3 and so on in the very beginning in stage number 1 you would have k 1 here and you would have E 0 and E 0 till day here. So let us consider stage 1 and let us derive some insights from stage 1 you have E 0 here E 0 till day there or E 0 Z and E 0 till day Z Z inverse as you will k 1 up there and k 1 down here E 1 Z here and E 1 till day Z there let us explicitly write down the expressions and let us join them together and take this to be the input. Now let us see what happens clearly E 1 Z would be equal to you know it is going to be E 0 Z plus k 1 Z inverse E 0 till day Z and E 1 till day Z is Z inverse E 0 till day Z plus k 1 E 0 Z but E 0 Z is equal to E 0 till day Z and therefore we could write down E 1 Z by E 0 Z and E 1 till day Z by E 0 till day Z. E 1 Z by E 0 till day Z by E 0 Z or E 0 till day Z it does not matter is 1 plus k 1 Z inverse and E 1 till day Z by E 0 Z k 1 plus Z inverse and let us call this A 1 Z and let us call this A 1 till day Z and we notice something very interesting as a relationship between A 1 and A 1 till day. A 1 till day is the coefficients of A 1 in reverse order. So here the coefficients of the successive powers of Z inverse are 1 and k 1. Here the coefficients of successive powers of Z inverse are k 1 and 1. We are in reverse order and that can be expressed formally in the language of Z transforms as follows. A 1 till day Z is A 1 Z inverse multiplied by Z inverse but you can easily verify. So A 1 Z inverse would be 1 plus k 1 Z and therefore Z inverse A 1 Z inverse would be Z inverse plus k 1 which is indeed A 1 till day Z to verify it. If we do indeed connect E 0 and E 0 till day together as we have done here we can now show the mathematical induction that this will always be the case. That means if we define A and till day to be the ratio of E n by E 0 mind you E 0 and E 0 and E 0 till day are connected. So you know put stages one after the other first stage, second stage, third stage but connect E 0 and E 0 till day together. So what I am saying is this. In the lattice structure put E 0 equal to E 0 till day as we have done here. And then consider the nth stage output E n and E n till day. Define A m Z to be E n Z by E 0 Z and A n till day Z to be E n till day Z by E 0 Z till day for till day and non till day for non till day. In here is that we can prove the mathematical induction but A n till day Z is Z raise the power minus n A n Z reverse or the coefficients of A n in reverse order. This is the language for saying the coefficients of A n till day Z are the coefficients of A n Z in the reverse order. We shall now prove this by mathematical induction. Now it is very simple. We have already completed the base system because we have proved it for capital N equal to 1. Let us now complete the inductive system. So let us write that down. We have completed the base system. So indeed A 1 till day Z is Z inverse A 1 Z inverse. Let the result be true and n can begin with 1 and therefore A n Z A n till day Z is indeed Z to the power minus n A n Z inverse. Let this be true. We shall show that it is true for n replaced by n plus 1. So indeed let us consider the relationships between A n plus 1 and A n. You see A n plus 1 Z is of course as you know E n Z plus Z inverse K n plus 1 E n till day Z. Whereupon the same relationships hold between the A's because you can divide by E 0 on both sides. So you get A n plus 1 Z is A n Z plus Z inverse K n plus 1 A n till day Z. But we can write down the expression for A n till day Z by mathematical induct by the assumption of mathematical induction, inductive assumption. So we can write this as A n Z here plus Z inverse K n plus 1 Z to the power minus n A n Z inverse. If we have got A n plus 1 entirely in terms of K n, there is no till day anywhere. And the same thing can be done to A n plus 1 till day. A n plus 1 till day Z is similarly equal to plus 1 A n Z plus Z inverse A n till day Z, which is of course K n plus 1 A n Z plus Z inverse times Z to the power minus n times A n Z inverse. That is a very simple matter. What we need to do is to compare the coefficient reversed form of A n with A sorry the coefficient reversed form of A n plus 1 with A n plus 1 till day. So let us write down, you see let us write down the two equations together to make matters easy. So what we are saying is let us write down these two equations. A n plus 1 Z is from here, this equation we will write it down explicitly. Essentially A n Z plus K n plus 1 Z to the power minus n plus 1 please note A n Z inverse and A n plus 1 till day Z from here is K n plus 1 A n Z plus Z to the power minus n plus 1 here. So K n plus 1 A n Z plus Z to the power minus n plus 1 A n Z inverse and now from here let us write down the coefficient reversed form. You see it is very easy to show that Z to the power minus n plus 1 A n plus 1 Z inverse would involve two steps. It would involve replacing Z by Z inverse. So you see this would become Z inverse here and this would become Z inverse inverse so Z. This would also get replaced by Z to the power n plus 1 instead of Z to the power minus n plus 1. And then you multiply by Z to the power minus n plus 1 there would be a factor of Z to the power minus n plus 1 coming in here and this Z to the power n plus 1 would go away with the Z to the power minus n plus 1. So it is very clear that this gives you Z to the power minus n plus 1 times A n Z plus K n plus 1 A n Z inverse. I am sorry. So it would be Z to the power minus n plus 1 A n Z inverse you know Z gets replaced by Z inverse here plus K n plus 1 this term has got cancelled and this term has become A n Z. Now compare this expression and this expression here they are exactly identical and therefore it is very clear that A n plus 1 till the Z is indeed Z to the power minus n plus 1 A n plus 1 Z inverse which means essentially this is the coefficient reversed form of A n plus 1. So induction complete. What is true for n has also been shown to be true for n plus 1. It is a mystery why this property has excited us so much. We will see that in the next lecture. Why is it that we are so excited about this coefficient reversal form? Anyway I mean I would certainly urge you to verify that this operation of replacing Z by Z inverse first and then multiplying by Z raised to the highest negative power equal to the degree does coefficient reversal. I urge you to verify it. Anyway what I will do is illustrate by one example before we conclude. So suppose we have a second degree term. So for example we have A 2 Z which is of the form some A 2 0 plus A 2 1 Z inverse plus A 2 2 Z to the power minus 2 then A 2 Z inverse times Z to the power minus 2 its degree 2 would become Z to the power minus 2 times A 2 0 plus A 2 1 Z plus A 2 2 Z squared which is very clearly Z to the power minus 2 A 2 0 plus Z inverse A 2 1 plus A 2 2. So it is the coefficient reversed form you see the coefficients in increasing powers of Z are A 2 2, A 2 1, A 2 0. Whereas they are A 2 0, A 2 1, A 2 2 here and you can see from this the mechanism of coefficient reversal how coefficient reversal occurs. The mechanism is clear from this example. This property of coefficient reversal will actually help us well one way to look at it is it. So it gives a very clear relationship between these two system functions A n and A n tilde and it is a relationship that we will use very fruitfully later to derive another structure which has feedback. You see the structure that we have in this lecture has no feedback. So we could use it to realize only FIR system functions. What we shall do in the next lecture is to investigate how we can use this structure to realize any FIR system function. That means you see what we have right now is a forward recursion but now we need a backward recursion. How do you go from A n plus 1 to A n? We have gone from A n to A n plus 1 but what we will have is the final structure and you want to peel off stage by stage to get the individual case. That is what we shall do in the next lecture. Thank you.