 Welcome to the 33rd lecture on the subject of digital signal processing and its applications. We had begun a discussion on realization in the last few lectures and in the previous lecture we had also begun the specific theme of design of lattice structures. We have also motivated the reason for the lattice structure design. The lattice structure is important because it is of course a uniformly repeated structure but more importantly it gives some insights into the stability or otherwise of a system when the parameters are calculated. And while the direct form or the cascade forms also have their own modularity, the lattice structure has some interesting properties that we shall now slowly develop. The lattice structure takes a while to understand. It is a little difficult to understand for the beginner. So it is worth reviewing a few ideas before we go further even though that might be some amount of repetition. So let us look at a few ideas that we have discussed in the previous lecture once again briefly that we put our discussion in perspective. We had said that we would take a typical stage of the following form in the lattice structure. We said we would take the nth stage and what I am drawing here is the signal flow graph just for the nth stage. And the overall fr lattice appeared like this. There was essentially an input point we will call it E0 and E0 till day if you like. You have the first stage and then you have the nth. For the first stage output you would have E1 till day and E1. At the nth stage output you would have En till day and En itself. And this can of course continue. This is the nth stage, this is how the nth stage looks. And this is all n stages together. What we need to do in the previous lecture was to obtain a recursive relation between the system function from E0 or E0 till day equivalently to En and En till day. And the recursion was on the system function itself. That means we expressed the system function at the n plus 1 stage in terms of the system function at the nth stage. That was what we did last time. And the motivation for doing that is to see certain properties that evolve in all these system functions. What we have seen in the previous lecture was that if we took the nth stage system function Amz defined by Emz divided by E0z and En till day z defined by En till day z divided by E0z. Then the recursion on the system function is as follows. Am plus 1 z is Amz Z inverse Km plus 1 Am till day z. And Am plus 1 till day z is Z inverse Km plus, I am sorry Z inverse Amz. Let us write that down. Let us rewrite this. Km plus 1 till day z is Z inverse Am till day z plus Km plus 1 Amz. So is that if we take the basis step here, Amz is A0z which is 1 of course plus Z inverse Km plus 1 A0 till day z and A1 till day z is Z inverse A0 till day z plus, yeah so of course M plus 1 is equal to 1 here. So K1 A0z and of course A0z is equal to A0 till day z obviously which is equal to 1 and therefore we have A1z is simply 1 plus K1z inverse. We have seen most of this the last time. And A1 till day z is Z inverse plus K1 and therefore we observe that A1 till day z is Z inverse A1z inverse which is easily verified because Z inverse A1z inverse is essentially Z inverse 1 plus K1z which is indeed K1 plus Z inverse which is A1 till day z. So we have verified that essentially there is a reversal of the order of coefficients. The coefficients in order of powers of Z inverse are 1 and K1 in A1z and in A1 till day they are K1 and 1. So essentially we are saying that the order of the coefficients in terms of powers of the inverse reverses when you go from A n to A n till day and we want to prove this by mathematical induction. Now we have of course gone through the details of that mathematical induction in the previous lecture. I shall not go through all of it again but we will just look at a few points to put our discussion in perspective because this is a very important topic and perhaps requires multiple attempts to comprehend completely. So what we saw then was that the inductive step could be carried out as follows. Note in that An plus 1z is Anz plus Kn plus 1 Z inverse A n till day z and An plus 1 till day z is Z inverse An till day z plus Kn plus 1 Anz and making the inductive assumption. An till day z is Z raised to the power minus n An Z inverse. Now what is this inductive assumption? This inductive assumption essentially is saying that you can obtain the coefficients of An till day by writing the coefficients of An in reverse order in terms of powers of the inverse. This is a mathematical way of saying this. Towards the end of the previous lecture we also explained why this is a mathematical way of saying that essentially you are reversing the order of the coefficients when you go from An to An till day. So making this inductive assumption and then using the two recursive steps we can then prove, we can prove An plus 1 till day z is Z to the power minus n plus 1 An plus 1 Z inverse which means that by inductive proof the coefficients of Z inverse are in reverse or in opposite orders or reverse orders between An and An till day. We have proved this. The reason why we have proved this is because if you learn An you should be able to write down An till day by a very clear algorithm and here we have a very clear algorithm. Once you know An, An till day is known. Now what we want to do next is to realize an FIR function. So suppose we have given an arbitrary FIR function H z H FIR z. We shall assume without loss of generality that z to the power of 0 has a coefficient of 1. I will explain why it is without loss of generality. You see suppose for example you have the following system function which violates this requirement. So suppose H FIR z is of the form z to the power minus 3 plus A1 z to the power minus 4 plus A2 z to the power minus 5 plus A3 z to the power minus 6. I can always extract z to the power minus 3 common from here and write 1 plus A1 z inverse plus A2 z to the power minus 2 plus A3 z to the power minus 3 is to be then taken. Take this. This is essentially 3 delays in cascade z to the power minus 3. So please you have 3 stages of that lattice where all the k's are 0. You see you can always write this. The first 3 steps has z inverse with 0 k's. You do not need to put the k's at all. Always put a cascade of 3 delays. In fact I do not even need to leave any blanks in between. So this realizes z to the power minus 3. So you can realize a few delays separately and the rest of it is then realized using the lattice structure. In any case the lattice structure is meant for giving insults into the part of the system function which we cannot obviously see. What we can obviously see that will require to be used in the lattice structure. It is the rest of it. The 1 plus 1 of this part is what we are interested in after all. We are trying to see what properties this part has. 1 plus a1 z inverse plus a2z to the power minus 2 plus a3. It is this part which is of interest. So at that part can be realized with the lattice structure. What is anyway trivial can be put separately. Now the next step what is called a synthetic or synthesis recursion. You see what we did so far was an analysis recursion. Analysis means you are moving forward from a0 to a1 to a2 to a3. That is analysis. So you will understand how to build a certain an. But now we want to build the other way. You see you have given the kns and you want to build an an. That is analysis. You have given the final an when you want to build a kn that is synthesis. So you want to construct a lattice structure that realizes a certain system function that is synthesis. Which means you are at the end now. So let us see. Now it is very clear that every time you add one lattice stage you are increasing the degree of an and an tilde by 1. In fact that is very easy to see. Let us go back to the expression. You see if you look at it in the recursive step it is very clear that a1 and a1 tilde of course have a degree 1. What happens to a suppose we assume that an and an tilde have a degree n. I do not need to write down this proof and just going over it orally. So you see let us assume by the inductive step the inductive assumption that an and an tilde have a degree n. Now this has a degree n. This has a degree n multiplying by the inverse makes a degree n plus 1. So you have this sum would be of degree n plus 1. So by inductive assumption by inductive step this has a degree n plus 1. Similarly this has a degree n. So this would have a degree n plus 1 and degree n plus 1 plus degree n would give you a degree n plus 1. And therefore both an plus 1 and an plus 1 tilde are going to be of degree n plus 1 that is quite clear. What is also clear is that the leading coefficient that means the coefficient of z to the power of 0 in the coefficient of z to the power of 0 in an is 1. Now we can see that again from the basis step. In the basis step a1 has a coefficient of 1 for z to the power 0. Let us assume the inductive step that an has for its leading coefficient that means the coefficient of z to the power of 0 equal to 1. Now of course you will notice that this term does not contribute any z to the power of 0 at all because there is a z inverse multiplying it. So all the terms have at least a power of 1 and z inverse and therefore the contribution to the z to the power 0 term comes only from this. So therefore the z to the power 0 term is carried as it is from an to an plus 1. And therefore the z to the power of 0 term is 1 always in all the an's. Now as a consequence it is very interesting to see that if you look at the coefficient of z to the power of n minus sorry z to the power minus n plus 1 the highest power of z or z inverse rather. So if you look at the highest power of z inverse in an plus 1 what would it be? You see it is very simple. In an plus 1 till day the leading term is k n plus 1 because this has no contribution to the constant term. The constant term comes only from here and the constant term is 1 in an z and therefore it is going to be k n plus 1 in an plus 1 till day z. So in an plus 1 till day z the constant term is k n plus 1. Now if the constant term is k n plus 1 in an plus 1 till day z then the highest power of z inverse will carry the coefficient k n plus 1 here because the coefficients are in reverse order. You see the coefficients of an plus 1 and an plus 1 till day are in reverse order. So the constant term in an plus 1 till day is the term carrying z to the power minus n plus 1 or the highest power of z inverse in an plus 1. Now what is the constant term in an plus 1 till day? This has no contribution to the constant term. So it is only this which has a contribution to the constant term and the constant term here is the constant term here multiplied by k n plus 1 and the constant term here is 1 and therefore the constant term in an plus 1 till day is k n plus 1 and therefore the term or the coefficient associated with the highest power of z inverse in an plus 1 is k n plus 1. Is that clear to everybody? Let us write down. This is a very important observation that we have made. So we have just proved z to the power of 0 carries a coefficient of 1 in am z for all z for all n z to the power minus n carries coefficient in am z at the end of the synthesis. That means suppose you indeed have ensured that the leading coefficient that is the power the z to the power of 0 carries the coefficient 1 then the coefficient associated with the highest power of z inverse is automatically the last lattice coefficient. So at least one part of the job is done for you. Now there is one thing that we have to keep in mind here. You see when we go backwards what we will need to do is to express a n and a n till day in terms of a n plus 1 and a n plus 1 till day. So we would need k n plus 1 to be known. So far you know actually k n plus 1 is a part of the synthesis. Now let us be clear. See all this while we have assumed that we knew the k n's and knowing the k n's we are calculating the a n's. Now we are going the other way. We know the last of the a n's. We know the final system function. We want to obtain the k n's which realize that system function. So we have to begin with the last one not the first. Right? So let us do that. Let us begin with the last. So in the synthesis recursion we begin with the last stage. Yes. The last a n will give us the system function. So the question is would the last a n give us the system function? Yes indeed. And the last a n till day. So the question is what happens to last a n till day. The last a n till day is an auxiliary function that we are going to obtain for some other purpose. So what we are going to do is we are going to use the a n explicitly and we are going to use a n till day implicitly in a recursion for some other reason. So it is an auxiliary quantity which we are getting for some other work. Right? We begin with the last stage. So we assume that h f i r z is equal to a n z and n is of course equal to the degree of h f i r. That is obvious. Now incidentally again that is issue about loss of generality. See in this again we have perhaps you know not so justifiably assume the coefficient of this to be 1. But even if it is not you know suppose the coefficient is a 0 here. You can always extract the a 0 here and divide everything by a 0. That is not really a problem. So even if this coefficient is not 1 you can always extract it common. So it is indeed without loss of generality that we can take h f i r to have the coefficient of z to the power 0 equal to 1 and in that case the degree of the 2 the degree n is equal to the degree of this f i r system function. And therefore k n is simply the coefficient of z raised to the power minus n in a n z which we know very well. Now this is essentially what we call the basis step in the synthesis because we have to begin with something. So here we have a basis the synthesis begins knowing the k n and now we have to obtain k n minus 1 k n minus 2 write down to k 1. So we need to go one step backward. So we need a backward recursion.