 A warm welcome to the 34th lecture on the subject of wavelets and multirate digital signal processing. This lecture intends to build upon the structure that we started discussing in the previous lecture. Let us recall in a few words what we were trying to do in the previous lecture. We were trying to build a modular structure to realize the orthogonal analysis filter bank. We noted that in building a modular structure, we would have an inductive approach, an approach where we would construct the smallest order orthogonal filter of length 2 and we demonstrated how the Haar analysis filter bank can be so constructed. And then to expand the length by 2 every time, we would need to introduce an expansion module, an inductive module. We introduced that inductive module and we proved by mathematical induction that you get an orthogonal filter bank of length 2 more every time we introduce one instance of that module. Now, this was constructive in the sense that we assumed that we were drawing a structure of this kind. We were putting together modules of this kind and we noted that we had the conjugate quadrature filter relationship or relationship in which the orthogonal filter bank conditions on the analysis filter side were obeyed. But what we wanted to do was to go the other way. Given an orthogonal filter bank, we needed to construct such a lattice structure which could realize the orthogonal filter bank. So, for example, if you had a Dobash filter of length 4, you would want to construct a lattice structure to realize it. And it was in that direction that we wrote down the recursions that govern the modular structure and we were trying to use those recursions to go one step back to peel off one layer as it were. Let us build on that peeling off process further today. Now, what we said was something like this. We said you have one modular stage like this here. You had the system functions h m z come up to here, h m tilde z come up to there. And assuming that we had taken the down sampler well beyond. So, the down sampler is at the end after all the modules. What appears next here is a z to the power minus 2. Following that, we have the m plus 1th lattice coefficient as it were which we called k m plus 1 here. And the negative of the same coefficient here minus k m plus 1. And this brought us to the next system function which we called h m plus 1 z. And h m plus 1 tilde z where we noted that the conjugate quadrature relation between h m and h m tilde is preserved in h m plus 1 and h m plus 1 tilde. What we mean by that is given that h m tilde z is z raised to the power minus 2 m minus 1. So, h m minus z inverse we have h m plus 1 tilde z is indeed z raised to the power minus 2 m plus 1 minus 1 h m plus 1 minus z inverse. So, the conjugate quadrature relation is carried. It is carried past the module that is what we are saying. Now, what we want to do is to go the other way. So, our construction construction of a lattice structure essentially means go the other way trying h m plus 1 and h m plus 1 tilde to h m and h m tilde. And h m plus now towards that objective we need to write down the recursive expression the inductive expression again that would give us a clue how to go backwards. So, let us write down the inductive relations once again inductive or recursive lattice relations take you from h m and h m tilde to h m plus 1 and h m plus 1 tilde. And the relations are h m plus 1 z is h m z plus k m plus 1 z to the power minus 2 h m tilde z. And h m plus 1 tilde z is z raised to the power minus 2 h m tilde z minus k m plus 1 h m z. Now, if we observe these two relations carefully and we wish to extract h m in terms of h m plus 1 and h m plus 1 tilde all that we need to do as we see is to cancel out this term. And that is easily done by multiplying this expression by minus k m plus 1. So, if you multiply this equation by minus k m plus 1 and add it to this term would vanish. What we are saying in effect is consider h m plus 1 z minus k m plus 1 h m plus 1 tilde z very clearly this would be equal to h m z plus k m plus 1 squared h m z plus k m plus 1 z raised to the minus 2 h m tilde z minus k m plus 1 z raised to the power minus 2 h m tilde z. And which of course is very easily seen to be 1 plus k m plus 1 squared h m z. So, we have eliminated h m tilde. So, we have an easy way now of obtaining h m from h m plus 1 and h m plus 1 tilde. Now, you know what we are doing here in terms of real computations or actual realization is starting from a higher order filter and going one step lower in the lattice. And what we have just shown is if you know the higher order filter you of course know its conjugate quadrature filter. Once you know h m plus 1 it is easy to construct h m plus 1 tilde all that you do is to replace z by minus z inverse in the argument and then multiply by z raised to the power a sufficient negative power. So, as to make it causal we shall illustrate this specifically for length 4 in a short while, but let us get the algebra complete first. So, what we are saying in effect is h m z is h m plus 1 z minus k m plus 1 h m plus 1 minus k m plus 1 minus k m plus 1 h m plus 1 tilde z divided by 1 plus k m plus 1 squared. Now, let us look at the validity of an expression like this validity means is this expression computable does it make sense. Now, of course this definitely makes sense there is no problem, but we do not know what k m plus 1 is we need to reason that out and that will also require little bit of reasoning. So, if I knew k m plus 1 this is in fact very easy to compute and so is the denominator there. So, all that we have to do is to reason out how to find k m plus 1 and in fact that we shall do by looking back at the forward recursion once again, but you know as far as this denominator is concerned here the validity of division by 1 plus k m plus 1 squared cannot be questioned as long as k m plus 1 is real. If k m plus 1 is real we have no problem of validity at all because this is definitely going to be a non negative strictly non zero quantity. So, all that we need to do is to obtain k m plus 1 how and that is done very easily by looking at the forward recursion. In fact, here again we will need to do a bit of inductive reasoning. So, let us write down the main step of the forward recursion once again h m plus 1 z is h m z plus k m plus 1 z raise the power minus 2 h m tilde z. Now, notice that h m plus 1 is going to be of length 2 into m plus 1 and these are going to be of length 2 into length 2 m. So, this factor of z raise the power of minus 2 is going to push this length 2 m filter 2 steps forward and that is how you are going to get a length of 2 into m plus 1. You know this pushing forward by 2 is what increases the length by 2. So, this is critical as you can see in increasing the length. Let us make a simple observation about the coefficient of z raise the power of 0. We shall show now inductively coefficient of z raise the power 0 in h m z is always 1 for all m. In fact, this is true in the basis step. You will recall that in the basis step the system functions were very simple. They were essentially 1 plus k 1 z inverse and minus k 1 plus z inverse and here of course, this was the coefficient of z raise the power of 0 in h 1 z if you please and therefore, it is true. Now, we shall use the inductive step to continue this proof. Let it be true for h m, m greater than equal to 1. Now, as far as h m plus 1 is concerned, h m plus 1 z is equal to is h m z minus k m or rather plus k m plus 1 z raise the power minus 2 h m tilde z. And let us go back a couple of steps in our reasoning when we wrote this equation. You know you can visualize the situation. Where would the z raise the power of 0 coefficient here come from? It would be a combination of the z raise the power of 0 coefficient here and if any in this term. But please note that in this term, you have a z raise the power of minus 2 common to all the terms in this expansion. So, this was of length 2 m and by multiplying by z raise the power minus 2, the lowest power of z inverse in this entire term is 2 now. So, you have no z raise the power of 0 term here. The only z raise the power of 0 term is here and that has been assumed by induction to be 1 and that is carried over here. So, it is this equation which carries over the z raise the power of 0 term from here to here making it 1 again. Let us write that down clearly. So, what we are saying is in this recursion, the z raise the power of 0 term is carried from here to here and of course, by inductive assumption if the z raise the power of 0 term is 1 here, it is also 1 there. So, this completes the inductive proof. So, now we know the coefficient of z raise the power of 0 is always going to be 1 in all the h m's. Now, it is very easy to see what k m plus 1 is. So, in fact, let us go back to the same equation that we used a minute ago. Let us look at this equation once again here. So, h m plus 1 is h m z plus k m plus 1 z raise the power minus 2 h m till day z. Now, let us understand what h m till day z is all about. So, h m till day z we have assumed in fact, we have shown inductively that it must be of the form z raise the power minus 2 m minus 1 h m minus z inverse and h m said we know what that looks like. So, h m z looks like this. It looks like 1 plus 1 let us say h 1 z to the power minus 1 plus h 2 z raise the power minus 2 plus h 2 m minus 1 z raise the power minus 2 m minus 1. Now, if you want to be very careful, you should superscribe this with m here to emphasize that this corresponds to the m th system function. So, I will do that to be careful. So, all we are saying is these are the impulse response coefficients or the impulse response points in h m and you know that the coefficient of z raise the power of 0 is 1. So, this is the form of the system function h m z. As a consequence, h m till day z is now going to be of the following form. You know h m till day z is first going to have z replaced by minus z inverse here. So, to do that we have 1 minus h 1 m z plus h 2 z raise the power minus 2 m minus 1 plus h 2 z raise the power h 2 m z squared and so on until we come to minus h 2 m minus 1 m z raise the power 2 m minus 1 and this whole thing is then multiplied by z raise the power minus 2 m minus 1. So, if you look at what is happening, these positive powers are going to be shadowed over by this negative power. So, for example, the h 2 m minus 1 m term is going to have z raise the power of 0 with it now and then you are going to have increasing negative powers of z as you come down and finally, this is going to correspond to z raise the power minus 2 m minus 1 right there. So, let us write the expansion again. So, h m till day z is going to have the form minus h 2 m minus 1 m plus and so on plus h 2 m z raise the power minus 2 m minus 1 plus 2 and as you keep going downwards then you have until you reach z raise the power minus 2 m minus 1 here. This is important and now making this observation, the coefficient of the highest and of course, odd negative power of z in h m till day is 1 and we carry that here. So, the coefficient of the highest power of z inverse in h m plus 1 is only going to come from here it cannot come from here. The highest negative power here is z raise the power minus 2 m minus 1, but since this is of length 2 into m plus 1 the highest negative power of z inverse here is going to be too larger than the highest negative power here and that can only come from here and in fact, as you can see that must be k m plus 1. So, what we are saying in effect is as a consequence of this let us write it down clearly as a consequence of h m plus 1 z is equal to h m z plus k m plus 1 z to the power minus 2 h m till day z and you know what the form of h m till day z is. This is what h m till day z looks like the z to the power minus 2 m plus 1 comes only from here the highest negative power of z and in fact, the coefficient of z raise the power minus 2 m plus 1 that is the highest negative power of z equal to k m plus 1. So, now we have a very simple way of obtaining k m plus 1 k m plus 1 is simply the coefficient of the highest negative power of z. So, once you have the final system function which you are trying to peel off stage by stage look at the coefficient of the highest negative power of z and that itself is the first lattice coefficient simple. So, once you know k m plus 1 you know how to go down descend one step once we know k m plus 1 we can peel off one module. In other words we know h m plus 1 we can construct h m plus 1 till day and then h m plus 1 z minus k m plus 1 h m plus 1 minus k m plus 1 minus k m plus 1 minus till day z divided by 1 plus k m plus 1 square gives h m z. So, notionally we have got h m z h m z was a system function after the previous module, but this is notional. What we need to worry about is does h m z really have a length of 2 less now instead of again trying to do the algebra generally here as we have been doing so far. So, this was the general algebra it will be easier for us to take a specific example and see how this works. So, let us take a length 4 orthogonal filter that length 4 filter could for example be the Dovash length 4 filter, but anyway we shall not put specifically the numerical values of the impulse response coefficients because we do not need to understand how this recursion works. We would find it much easier to just carry out the recursion and demonstrate that it is feasible. So, let us take an example length 4 filter. So, it would have a form 1 plus h 1 z inverse plus h 2 z raised to minus 2 plus h 2 z raised to minus 3. So, we wish to peel off one stage. In other words what we are saying is we have this situation we have two stages of the lattice here. So, we know the system function here and therefore, also here known at this point and we wish to obtain k 2 and k 1 that is what we are saying. So, the problem is obtain k 2 and k 1. Now, what we just said makes it very easy for us to obtain k 2 given the length 4 filter 1 plus h 1 z inverse plus h 2 z raised to the power minus 2 plus h 3 z raised to the power of minus 3. What we discussed a few minutes before this namely that the most negative power of z reveals k to us. So, this essentially is k 2 and having noted that we can obtain one step lower. So, now we can construct if we call this h z or let us say h 2 z up to the second stage then we can also construct h 2 till day z. And h 2 till day z is going to be very simple it is going to be z raised to the power minus 3 h 2 minus z inverse which is z raised to power minus 3 1 minus h 1 z plus h 2 z squared minus h 3 z raised to the power of minus the power 3. So, h 2 till day z can be simplified it is minus h 3 plus h 2 z inverse minus h 1 z raised to the power minus 2 plus z raised to the power minus 3. So, all that we now need to do is to consider the numerator of notional h 1 z. That numerator is going to be h 1 z notionally as you remember was h 2 z minus k 2 h 2 till day z divided by 1 plus k 2 the whole squared. And if you only care to expand this. So, you know to expand the numerator it will be easier for me to write down the expansion in two lines one line for h 2 z and one line for minus k 2 h 2 till day z. So, h 2 z minus k 2 h 2 till day z and noting that k 2 is essentially h 3 we have 1 plus h 1 z inverse plus h 2 z raised to the power minus 2 plus h 3 z raised to the power minus 3 minus k 2 h 2 till day z which is minus h 3 z raised to the power minus 3 plus h 1 h 3 z raised to the power minus 2 and nothing else here. So, of course, we will just completed it is minus h 3 h 2 z raised to the power minus 1 and of course, plus h 3 squared here. Now, this is very interesting let us look at this expression carefully. Let us look at the coefficient of z raised to the power minus 3 this anyway becomes 0 identically. What is interesting is this the coefficient of z raised to the power minus 2 it is h 2 plus h 1 h 3, but let us go back to the basic requirement that we had on an orthogonal filter. Remember the whole idea in constructing the Dabash filter for example, was that the impulse response was orthogonal to its even translates. Let us put that condition down again and see what it gives us. So, 1 h 1 h 2 h 3 is orthogonal to 1 h 1 h 2 h 3 and that means h 2 into 1 plus h 3 into h 1 is 0. Now, go back to the coefficient of z raised to the power of minus 2 here that is h 2 plus h 1 h 3 low and behold that is the same as this and it is identically 0 on account of the orthogonality to even translates. So, let us make a note of that. So, this is 0 because of this orthogonality to even translates. So, what is left now remaining is 1 plus h 3 squared plus h 1 minus h 2 h 3 times z raised to the power minus 1. So, we are in a very good situation as far as h 2 z minus k 2 h 2 till there is concerned the length has gone down by 2 as we indeed wanted it to. Accordingly h 1 z is now going to be 1 plus h 3 squared plus h 1 minus h 2 h 3 into z inverse divided by 1 plus h 3 squared k 2 squared remember here and that is very simple that is 1 plus h 1 minus h 2 h 3 divided by 1 plus h 3 squared times z inverse and low and behold this also reveals to us k 1 directly. So, we have completed the lattice. So, this is the construction and as you can see for the dobash case this is going to work definitely all that it required is orthogonality to even translates nothing else was needed. Because of the orthogonality to even translates when we carried out the numerator part of the peeling process the length decreased by 2. Now, you know one must understand where there is something unusual here in going to h m of that matter h m till day from h m plus 1 of that matter h m plus 1 till day there was just a little bit of solution of 2 equations involved that could have been solved for any h m plus 1 and h m plus 1 till there is nothing very special about the form of h m or h m till day for that matter in writing down that downward step. However, what was important is that when you went from h m plus 1 to h m you needed the length to go down not by 1, but by 2 and this decrease of length by 2 is what brought in the requirement of orthogonality to even translates. So, it is only because of the orthogonality to even translates that the length decreased by 2 in the numerator part of the expansion and of course, once you were assured that the coefficient k which you had peeled off how did you peel off the coefficient k by looking at the coefficient of the highest negative power of z in h m plus 1. So, the coefficient of the highest negative power of z in h m plus 1 gave us yielded or in fact, clearly revealed the corresponding k of the outermost lattice stage. So, now, we can see how to peel off stage by stage we have done it specifically for a 2 stage lattice and it is not difficult to extend this to a 3 stage or a multi stage lattice beyond 3 stages. Now, a few remarks are in order here you know what we have done now is actually to demonstrate with an example that when you carry out this backward recursion let me now write down the more general backward recursion again let us make these observations. The backward recursion h m z is h m plus 1 z minus k m plus 1 h m plus 1 tilde z divided by 1 plus k m plus 1 squared is effected by the following steps. Step number 1 k m plus 1 is the coefficient of highest negative power of z inverse in h m plus 1 as long as this coefficient is real the denominator which is 1 plus k m plus 1 squared poses no problem. What I mean by that is a denominator can pose a problem if it is 0 you cannot divide by 0. So, you are assured that this is not going to be 0 if k m plus 1 is real or in other words the coefficient of the highest power of z inverse is real you are guaranteed this denominator there is no there is not going to be any division by 0. Having made that observation now the third observation is in the numerator namely h m plus 1 z minus k m plus 1 h m plus 1 tilde z. There is a denominator in the denominator length reduction of 2 and the justification you see reduction by 1 is easy. Essentially if you look at this numerator the coefficient of the highest power of z inverse here is k m plus 1 and the coefficient of the highest power of z inverse here without this would have been 1, but with this it is minus k m plus 1. So, that would cancel out. So, essentially reduction by 1 is easy it occurs by cancellation of highest power of z inverse, but the additional one step reduction additional reduction in degree by 1 is attributed to orthogonal to even translates. This is a very important observation this is where the orthogonality of the filter bank comes into play where it not an orthogonal filter bank that we were talking about we would of course, have trouble in getting this additional reduction of degree. Now finally, we need to complete this lecture by putting in a variant and that variant is essentially to transpose the structure that we have constructed. In other words how could we use the lattice structure that we have constructed to build a lattice structure for the synthesis side. So, let us complete this lecture by discussing this variant the synthesis variant. Now it is very easy to construct a synthesis variant essentially it is the transpose of the analysis lattice and what I shall do is to construct the synthesis variant first for a length 2 and then for a length 4 and that could illustrate how we could do it in general. So, length 2 synthesis lattice is essentially the transpose of this and you can visualize what the transpose will look like. Let us just go over it orally first and then draw it we would have the transpose with all these arrows reversed here. You would have up some plus at this point these arrows reversed the z inverse as it is this now becoming a summing point instead of a branching point and then resulting in the output here. So, let us draw the transpose there we are this is the transpose of a length 2 stage. Now all that we need to do to complete this discussion is to show the transpose of the inductive module. So, the inductive analysis module was essentially this of course this is after the down sampler let us say k m plus 1 here and minus k m plus 1 there and therefore, the corresponding synthesis inductive module would look like this. Once we have the inductive modules and the basis module it is easy to construct the complete lattice on the synthesis side and now I leave it as an exercise for the class at the end of this lecture to work out the same kinds of recursions as we have done for the analysis side to construct the synthesis filter bank. Leaving that exercise to you we conclude the lecture here to observe that we can build a beautifully computational efficient structure called the lattice to realize an orthogonal analysis and now synthesis filter bank. Thank you.