 we need to invert the following. You see a n plus 1 let us write down the recursion but this time let us note that we already have a relation between a n and a n till day. So a n plus 1 z is a n z plus k n plus 1 z inverse z to the power minus n a n z inverse. Let us write that down straight away and similarly a n plus 1 till day z is z inverse times z raised to the power minus n a n z inverse plus k n plus 1 a n z is to get the a n from the a n plus 1 now. You see so we need to invert this. So that is very easy to do. I just need let me call this relationship 1 and let me call this relationship 2. All that I need to do is to cut off this term really you know I can try and get rid of the a n z inverse term that is very easy to do. I can just subtract k n plus 1 times this from this. All that I need to do is to operate minus k n plus 1 times 2 and that gives me a very simple and elegant relationship. It says a n plus 1 z minus k n plus 1 a n plus 1 till day z is a n z minus k n plus 1 squared a n z as simple as that. Now you see this is of course very easy to write 1 minus k n plus 1 squared a n z. So now we have a n z in terms of a n plus 1. In fact we know how to construct a n plus 1 till day z from a n plus 1 z. We know how to do that by the inductive relation. So in fact what this says is that a n z is then a n plus 1 z minus k n plus 1 z to the power minus n plus 1 a n plus 1 z inverse divided by 1 minus k n plus 1 squared. Now the most important step completed the backward step from a n plus 1 you can go one step back. Now we can think of this as peeling off one stage. You will want to visualize the lattice structure like a piece of cabbage where you have peel off the peel off the peel. Each stage is like a peel. So what we have done is to peel off one layer of that cabbage and you see the inner layer now. Now you can peel off now you see this is recursive. So you know how to go from a n plus 1 to a n and the same approach can be used to go from a n to a n minus 1 and you can keep doing this until you reach a 1 and once you reach a 1 of course you are done. Now in this process the properties of a n would be maintained because if you look at it again here you see it is very clear that see the whole idea was that we would need to maintain the property that the coefficient of z to the power of 0 is 1 and so on. Now that would continue to be maintained in this recursion. I leave it to you to verify that. But one word of caution there is only one situation in which we will have trouble here and that relates to when this denominator becomes degenerate. So there is degeneracy when k n plus 1 equal to 0. That means in this lattice recursion at any stage if you find that this k turns out to be 1 then you have a degeneracy you cannot realize it with the lattice structure. So I am sorry I am sorry yes I meant to say k n plus 1 I meant to say 1 minus k n plus 1 squared is equal to 0 yeah that is correct. k n plus 1 in fact I would write I would prefer to say k n plus 1 squared is equal to 1 and of course we are assuming that the coefficients are all real ks are real. So it could be either plus or minus 1 right. So plus or minus 1 lattice parameter this is called the k n is called the lattice parameter. Let us give the let us take note of this term k n is called the lattice parameter. So degeneracy results when the lattice parameter takes the value plus or minus 1 and degeneracy means that that system function cannot be realized with the lattice. In fact it is of course an important question why a degeneracy takes place and why these lattice parameters are so interesting and this just looks like any other structure what is so special about this. We will now put down a very important theorem which will explain why this is so important. I shall not prove this theorem but I shall state it. The lattice parameters indicate the location of the roots of H FIRZ with respect to the unit circle in the following specific way. The roots of H FIRZ within the unit circle that means all of modulus less than unity. Now not strictly less than unity if and only if all the lattice parameters corresponding to H FIR are less than 1 in magnitude that means if the system function H FIRZ has all its lattice parameters less than 1 in magnitude you are guaranteed that it has its roots inside the unit circle and vice versa. If you require that H FIRZ have all its roots inside the unit circle then all the lattice parameters must be less than 1 in magnitude. Now why is this important in fact not so much for an FIR function. It is important for the denominator in a rational function that means what we want to do is to exploit this property to study the denominator of a rational system function and that is the next step that we would like to take and that is the reason why as somebody asked why are we creating that auxiliary function a until day. The auxiliary function a until day is being created to allow us to exploit this property to realize a rational system function in general. So we are actually going to use these k's to realize the denominator and the numerator would be realized with some other tap of coefficients and then you get inside into the stability of the rational system function by looking at the k's. Now what does it mean about the k equal to 1 k so what k equal or k squared equal to 1 k's it means there is trouble. If you find that any of the k's becomes plus minus 1 then you have a problem with the solutions then if and only if relation. So you know if you are trying to look at so you know what you could I mean irrespective of whether you are realizing a rational system function. What this tells you is that if I am given the denominator polynomial in a rational system function and of course I have told you that in a causal rational system function you can always write the denominator with a leading 1 in a causal I repeat the causal rational system function you are always in a position to write the denominator with a leading 1 the coefficient of z to the power of 0 is equal to 1. What is to do that take the denominator as it is think of it as a system function and realize it with the lattice structure and then if you want to check whether the system is stable or not all the solutions must of course lie within the unit circle and that implies that all the lattice parameters must be strictly less than 1 in magnitude. So it is a very simple test for stability of a rational system function let us make a note of that. So this yes there is a question. What happens if the magnitude is greater than 1. The question is what happens if the magnitude is greater than 1 or in fact for that matter what happens if the magnitude is equal to 1. Well all that we are saying at the moment is that if the magnitude if all the magnitudes are not less than 1 strictly you are sure that there is a root either on the unit circle or outside that is all that we can say at the moment. We are not going further at the moment right. So there is a 1 to 1 I mean there is stability implies and is implied by or you know what I mean is if you want to check that all the roots of a polynomial and Z inverse are within the unit circle then it is necessary and sufficient that all the corresponding lattice parameters be less than 1 in magnitude. If that condition is violated you are sure that there is a violation also of the roots being inside the unit circle. Now in what way we do not know whether the roots are on the unit circle or the roots are also outside that we are not going into at the moment. In fact we have not even proved this. This proof is a little involved. Time permitting we will see if we can give a small proof. I would also like to offer it to you as a challenge but it is a rather difficult challenge. Well actually that the I can tell you if some of you are really interested in taking up the challenge and trying to prove this the central idea in the challenge is to look at what happens when you create an all-pass like function as you have here. You know the relation between a n and a n till day and the recursive relation between a n and a n plus 1 is at the heart of this proof. You know anyway I do not want to spend too much of time on that. Save to say that the proof is a little involved and we shall take it up if time permits but we must know the consequences. The theorem should be understood very. Theorem is not too difficult to understand. It says that necessity and sufficiency is essentially it states the necessity and sufficiency of all the lattice parameters being less than 1 in magnitude for all the roots of that polynomial to be inside the unit circle. Now what why is this useful that is what we are going to do next. What we are going to do next is to assume then that we have a rational causal system function as follows. As usual it is of the form summation n going from 0 to capital M be m z raised to the power minus m divided by 1 plus or minus whatever you like does not matter. L equal to 1 to m L z raised to the power minus L. We want to realize with a lattice. Now obviously you cannot do it as it is. So far a lattice only allows us to realize a FIR system function and the real use of the lattice is in studying this. So what we need to do is to make a little change in the equations or the recursive equations of the lattice. We could go back let us go back and look at those recursive recursions once again. So in fact maybe we do not even need to do that. Let us write them right from fundamentals. So we will not even write the system functions. Now we will write the sequences right from the beginning. We will go back to the very first pair of equations that we wrote in terms of sequences on the lattice. So we had En plus 1 z is En z plus Kn plus 1 z inverse En tilde z and En plus 1 tilde z is z inverse En z En tilde z I am sorry plus Kn plus 1 yeah En z. Now all that we do is to make a little rearrangement of this. So we just rearrange it and rewrite this as En z is En plus 1 z minus Kn plus 1 z inverse En tilde z. So it is an interesting situation that we have here. We have En plus 1 tilde written in terms of En tilde and En and we have En again written in terms of En plus 1 and En tilde. Then let us write down the structure that this implies. Now please note that we have not changed the basic equations. The basic equations are the same. So therefore the basic relationships remain the same. The basic relationships do not change. But the structure changes and the structure would look like this. You see what we will do is we will now draw the structure the other way. So we will move this way from E0 to En or En tilde. So we assume that somewhere here you have an E0 and E0 tilde and you are moving this way. So we write one stage. So we have En tilde here please not En and of course En there and go back to the equation. We have an En tilde z multiplied by z inverse. So there we are. What is this equation say? This equation says z inverse times En tilde needs to be multiplied by minus K n plus 1 and add it to En plus 1. In fact let us make this En plus 1 here for convenience. Oh well I am sorry. I think this is alright. So you need to multiply this by K n plus 1 rather minus K n plus 1. What about this equation here? En plus 1 tilde z. So you know you have let us keep forming the structure. So it is a little complicated. You have En plus 1 tilde being formed here. En plus 1 tilde is formed with a z inverse En tilde z and K n plus 1 times En z. So you need a K n from here and this as it is. You see En plus 1 tilde z is z inverse En tilde z. Z inverse En tilde z is essentially this plus K n plus 1 En z. So you need a K n plus 1 to come from there. Let me write it like this. You know let us remove this branch for the moment. Let us take this later. Let us draw only one pair of branches to make it look much neater. You have En tilde. You have a z inverse. We realize this equation first. So we add this to this multiplied by K n plus 1. So what we have done is to realize this equation here. You see the second equation is En plus 1 minus K n plus 1 times z inverse En tilde. You see so that is a funny situation to be in. You have an En plus 1 here. You need to take En plus 1 and subtract from it K n plus 1 times z inverse and sorry yeah z inverse times En tilde. That is you need to realize this equation in the second branch. Now how would you get that? You would need to take En plus 1 as it is and subtract from it K n plus 1 times z inverse En tilde. Now you are getting z inverse En tilde here. Multiply that by minus K n plus 1 and add it to En plus 1. And this is what should give us En. In fact now we have a problem. You see now this is the kind of structure that we expect. We need to see more of this. We will take this up again in the next lecture. We will sort of clean up the structure little more. So we have to do a little bit of shifting here. We will have to rearrange the structure to realize this set of equations. But the philosophy is the following. The equations are the same. The relationships are the same. But now we are reinterpreting the equations to give us an IR system function, rational system function. We need to spend more time on this and therefore we will continue this in the next lecture by repeating some steps. Thank you.