 A warm welcome to the 27th lecture on the subject of wavelets and multirate digital signal processing. In the previous lecture, we had completed or almost completed the proof of the theorem of multiresolution analysis. So, in a certain sense, we are at a cross point, a juncture, where we need to explore more avenues. We have understood a whole approach to building multiresolution analysis. We have talked about orthogonal filter banks in great depth. We have also established the connection between the continuous time functions phi t and psi t and the filter banks. And we have brought out a proof that given the axioms of the multiresolution analysis, we are guaranteed the existence of a wavelet psi t, whose dietic dilates and translates can span the whole of L 2 R. And we did that by essentially considering one peel, we looked at one incremental subspace and we showed that that incremental subspace namely W 0 could be spanned by the translates of a single function psi t. And in fact, the function psi t was a linear combination of the basis elements in V 1 as expected. And we could also characterize the coefficients of the linear combination. So, in that sense, the proof that we had in the previous lecture was constructive. Now, the aim of the lecture today is first to complete a few details in the proof that we attempted the last time and to take further by introducing variance. So, what we intend to do today is first to begin with a few details of completion of the proof and that would in fact lead us to the possibilities of varying the notion of multiresolution analysis in ways that will make our time frequency possibilities richer. So, to take the first task that is to complete a few details of the proof of the theorem of multiresolution analysis. The first thing that we need to complete is to prove that psi t is orthogonal to phi t and its translates. We have already shown psi t belongs to V 1 of course and can be expanded as follows. psi t is summation on all integer n g n phi 2 t minus n, where g n is the impulse response of the analysis high pass filter. That is interesting. Expressing phi t in terms of the basis of V 1 involve the coefficients of the low pass analysis filter and expanding the wavelet in terms of the basis of V 1 involve the impulse response of the analysis high pass filter. In fact, just to complete an example let us look at the Haar case. In the Haar case the high pass impulse response is essentially 1 and minus 1 and as expected the Haar wavelet is indeed 1 times phi 2 t minus 1 times phi 2 t minus 1 as an example. Now, let us make an observation in general here. So, what we are saying in effect is if we take g n essentially to correspond to the inverse z transform of z raise the power minus l minus 1 h of minus z inverse, where h of z is the analysis low pass filter. Then what we wish to do is essentially to analyze the following dot product. The dot product of psi t with phi t minus m for any integer m and this is easy to do. Indeed psi t by itself is summation on n g n phi 2 t minus m phi t minus m is simply summation on n and here again I imply summation on all integer n. So, as so as not to confuse these two indices I shall write n 1 here. So, n 1 h n 1 phi 2 t minus 2 m minus n 1 taking a q from expanding phi t minus m in the basis of the high pass basis of v 1 as we can do where upon this dot product is essentially a summation on n a summation on n 1 g n h n 1 complex conjugate times the following dot product phi 2 t minus n minus 2 m minus n 1. Now, once again without going through the whole derivation again I might observe that this dot product is 0 save for the case when n is equal to 2 m plus n 1 and in fact that means that we can drop one of these two summations. We could if we like drop the summation on n what we are saying essentially is that phi 2 t minus n dot product with phi 2 t minus 2 m minus n 1 is half delta n minus 2 m plus n 1 and therefore, we can drop the summation on n and leave summation on n 1 g. Now, you know this is an impulse a discrete impulse which has a non zero value only when n is equal to 2 m plus n 1. So, in place of n we substitute 2 m plus n 1 of course, h n 1 complex conjugate and a factor of half which is not a terribly serious matter at all. Essentially, this is the cross correlation of the sequences g and h evaluated at 2 m for all integers m looks familiar does not it we have been doing this frequently. Now, one can again evaluate the cross correlation by going into the z domain. So, obviously the z transform of the cross correlation can be seen to be essentially g of z times h of z inverse and we know what g of z is g of z is essentially the system function of the analysis high pass filter which is z raise the power minus l minus 1 h minus z inverse and therefore, we have the cross correlation z transform essentially becomes z raise the power minus l minus 1 h minus z inverse h minus z inverse times h z inverse. Now, for the moment let us simplify matters assume the impulse response is real. We can of course, generalize all these results to when the impulse response is complex, but that is not the critical issue here. Let us not get confused by the complex conjugate every time. Anyway, now all that we need to do is to look at the values of the cross correlation at even locations and we know how to do that. If the z transform of the cross correlation is some function of z, we take the same function of z with z replaced by minus z at the two together and see what happens. That tells us about the values of the sequence at even locations. Let us do it for this cross correlation sequence. After that z raise the power minus l minus 1 h of minus z inverse times h of z inverse plus the same thing with z replaced by minus z is as follows. It is essentially z raise the power minus l minus 1 h minus z inverse h z inverse plus minus 1 raise the power l minus 1 z raise the power minus l minus 1 h z inverse times h of minus z inverse. This is very interesting. Remember l is even. We have agreed that our orthogonal filter bank is going to have an even number of samples in the impulse responses. If it does not, then we are going to have a problem in the requirement of orthogonality to even shifts. So, take for example the Doberts series. Obviously, their impulse responses are of even length. In fact, it is not too difficult to show that even if you assume an odd length, you will find the last sample must be 0 for the requirement that the impulse response be orthogonal to its even shifts. Anyway, coming back to this, since l is even, l minus 1 is odd and therefore minus 1 raise the power l minus 1 is minus 1. As you can see, this term and this term are exactly identical if you just interchange the order of the products. Therefore, these two cancel. This is identically 0. Therefore, the cross correlation of g and h is 0 for all 2 m and that amounts to saying that the dot product of psi t and phi t minus m is equal to 0 for all integer m, which is exactly what we had set out to prove. So, we have completed that little detail and in completing that detail, we have also brought out an important issue, namely the intimate connection between the z domain representation of the filters and orthogonality of the functions. We had seen this implicitly in the proof at points. In fact, now to take this discussion a little further and to give you a feel of the connection in a slightly greater depth, let me again take the issue of orthogonality of phi t with its own integer translates. How does that manifest on to the behavior of the autocorrelation of the low pass filter impulse response? So, in fact, if we look at the function phi t and its integer translates, consider the dot product of phi t and phi t minus m. We can see that it can use a similar set of steps as we have just done. So, I would not repeat all the steps that this is equal to summation on n h 2 m plus n h bar n and then of course, you have you know you have a factor of half there. So, it is essentially the cross or the autocorrelation of the impulse response of the low pass filter. Now, once again we are evaluating the impulse response at only even locations. So, let me first write down the z transform of the impulse response. You know you will remember the z transform of the autocorrelation of the impulse response would essentially be h of z times h of z inverse. What we are asking is what is h z h z inverse plus h minus z h minus z inverse. This is essentially what would tell us about the values of the autocorrelation at the even locations. Now, we know the answer to this by the very design of an orthogonal filter bank. This is essentially a constant. So, the design equations for an orthogonal filter bank ensure that this quantity h z h z inverse plus h minus z h minus z inverse is a constant and therefore, the autocorrelation is 0 at all even locations. Let us call them 2 m except for m equal to 0. That is why you get that constant non-zero constant. So, there is a very beautiful observation we have made. The so called power complementary property that is also a consequence of this essentially manifests as the orthogonality of the impulse response to its own even translates and that also becomes a part of establishing the orthogonality of phi t to its own integer translates. So, they are all related. Now, this again brings out the intimate connection between the autocorrelation behavior of the filters in the filter bank and the orthogonality requirements that we insist upon in the underlying continuous functions that are generated on iteration of the filters in the filter bank as we have done before. You know how to iterate the impulse responses, convolution of the impulse responses to go towards phi t and psi t. Now, what are the variants that we can introduce at this point? You see when we look carefully at these equations, I mean we put them all down together like this and when we look at them from these perspectives, we see that there are many things which we have not done to date which can we can probably explore and let us put them down one by one. The first thing we need to explore is must we have essentially the same analysis and synthesis filters. Of course, this leads to an orthogonal filter bank, but then you know remember when we talk when we move from the continuous wavelet transform towards the discrete wavelet transform going through one step where we discretized the scale parameter logarithmically, we had made a remark. We had come to a point where we said there are two options. Either have different wavelets psi and psi till day on the analysis and synthesis side and build different analysis and synthesis filter banks and get perfect reconstruction or use the same wavelet on the analysis and synthesis side, but that so called quote unquote same wavelet would be different from the wavelet from which we started. In case the discretization of the scale parameter was such that the sum of dilated spectra did not come to a constant for all frequencies, but lay between two positive bounds, we could bring in a new wavelet which gave us an orthogonal analysis and synthesis filter bank. The wavelet would be the same psi double till day, but different from the original wavelet psi from where we started. So, there were two options. Now, we must explore the same two options in the context of discrete filter banks. Establishing the connection between the continuous case and the discrete filter bank is an involved exercise, but taking inspiration from the continuous case from the continuous wavelet transform and realizing that there are these possibilities to be more general namely that you could have different analysis and synthesis filters and come up with a variant of the concept of multi resolution analysis is certainly something that we should explore in depth. And in fact that needs to be explore in depth for another reason. One of the most recent data compression standards actually employs a filter bank with that perspective. Let me say a little more about this. You see wavelets have been rather successful in the area of data compression. In fact, the central idea in wavelet based representation is that you can represent both global and local information efficiently with a few number of with a few coefficients. So, you could take transient information or short lived information and represented with few coefficients in the details and long lived information or you know information spread all over the interval over which the signal last and also represented with few coefficients in the approximation part. And therefore, wavelets are an attractive proposition for data compression. Now, using this fact the joint photographic experts group or GPEG as it is called for short came up with what are called the GPEG 2000 standards. 2000 refers roughly to the year around which these standards were finalized. So, GPEG 2000 standards for data compression actually employ what are called bi orthogonal filter banks. And in the lecture today we shall look at these bi orthogonal filter banks in some depth. So, bi orthogonal filter banks are filter banks where the analysis and the synthesis filters are not quite the same. In other words you do not just have one analysis low pass filter from which everything else is derived by small variations of replacing z by minus z or z by z inverse and so on. There are essentially in a in fact in the GPEG 2000 standards. There are essentially two filters and the other two are derived out of them. So, we come to that in slightly greater depth in a few minutes. But before that let us look at some other variants that are possible. Now, so far we have been looking at finite impulse response filters. Do the filters in the filter bank have to be finite impulse response? Well obviously the answer is no. One does not have to have a finite impulse response which one could certainly conceive of an H z or for that matter a G z I mean H 0, H 1, G 0, G 1 where the sequence underlying the z transform is infinite in length as an infinite number of non-zero samples nothing stops us from considering such H z and G z. So, we have to admit that possibility and actually when we look at the 5 3 filter bank as it is called in the JPEG 2000 standards we will see that if I insist on getting an orthogonal filter bank solution for this kind of a paradigm we need to go to an infinite length impulse response filter bank. We shall build that up in a subsequent lecture. So, this is the second variant that we wish to explore must be restrict ourselves to finite impulse response filter banks. The third variant which we need to explore is as follows must we always iterate on the low pass branch and obviously again the answer is no. In principle nothing stops us from putting the whole filter bank back on the high pass branch let me put before you what I am saying graphically. What I am saying is this the discrete wavelet transform looks like this essentially we have a low pass filter and a high pass filter where the finite or infinite length that is not the issue followed by a down sample and we keep iterating here. So, we put the low pass filter here and the high pass filter there and down sample once again and keep doing this. Now, what about this can we put this filter bank here nothing stops us in principle. In fact it is obvious that if you were to take one stage of this namely if I were to do the following I were to take the input I were to subject it to the action of a low pass and high pass filter followed by down sampling as usual followed by iteration here LPF HPF down by 2 and down by 2 there. But also do the same here I could get back here by using the corresponding synthesis filter bank for this and I could get back here by using the corresponding synthesis filter bank for this and once I get back here and here. Now, please remember of course I would get back with a constant multiplier and a delay, but that is not such a serious issue I mean I can always take both the constant and the delay across the down sampler and I could actually absorb it with these filters here and finally, I could in principle reconstruct this once I have got here. So, whatever argument allowed me to decompose on the low pass branch every time can also allow me to decompose on the high pass branch. So, there is a new possibility that has opened up suppose I were to conceive of decomposing the high pass output what does it imply for the underlying functions and that would lead us to the idea of what is called the wave packet transfer. The central idea in the wave packet transform comes from iterating on the high pass branch as well not only the high pass branch, but allowing for iteration on the high pass branch also that would lead us to the idea of the wave packet transform. So, this is a brief introduction to the variance of the idea of multi resolution analysis that we wish to explore one by one. Now, let us begin with the first of the variance namely the filter bank used in JPEG 2000 let us not be too general. Let us consider one particular filter bank which is used in JPEG 2000 where the filters are of different lengths the analysis filters are of different lengths and that would lead us to the idea of what is called a bi orthogonal filter bank or a filter bank where we do not quite have orthogonal scaling functions or scaling functions which are orthogonal to its to their translates. Now, you know how the inspiration for the 5-3 filter bank in the context of JPEG 2000 could be seen to be the following there are various ways to explain the inspiration, but one of the inspirations can be seen to be the following. Let us begin with the Haar case where we have the dilation equation for phi t given as follows phi of t. Now, you know just to ease our notation we shall call the corresponding scaling function for the Haar MRA as phi 0 t we will understand why the 0 in a short while phi 0 t is phi 0 2 t plus phi 0 2 t minus 1 and you will recall that this translates graphically to the following this is phi 0 t this is phi 0 2 t and this is phi 0 2 t minus 1 this explains what we are saying graphically. Now, suppose we convolve this equation with itself. So, we write the equation down twice and convolve it with itself what would happen? Now, you know the central idea is what happens when you convolve say a term like phi 0 2 t with a term phi 0 2 t minus 1. In other words we are asking the following general question given h t convolved with g t to be r t what is h t with g a t plus b convolved with g a t plus c please note a the same a is used here b and c can be different, but the same scaling factor is used a of course belongs the set of real numbers same scaling that should be emphasized and a is real. In fact, let us evaluate it indeed h a t plus b convolved with g a t plus c as we desire here is essentially integral h a lambda plus b g a t minus lambda plus c d lambda integrated overall lambda and we use the standard principle of substitution of variable. So, let a lambda plus b be another variable let us call it gamma. Now, there are two possibilities a is of course not equal to 0 and a is real. So, when a is greater than 0 d gamma is of course always a d lambda and lambda going from minus to plus infinity translates to gamma also going from minus to plus infinity on the other hand when a is negative of course, d gamma is as usual a d lambda, but lambda going from minus to plus infinity translates to gamma going from plus infinity to minus infinity. However, if you look back at the limits of the integral here. So, you know when a is positive d lambda is going you are going to have d gamma as a d lambda. So, you know you are going to have a factor of a coming here in general a 1 by a. So, here d lambda is going to be d gamma by a. So, 1 by a is going to be positive if a is positive 1 by a is going to be negative if a is negative when a is negative these limits are reversed when a is positive the limits are where they are. So, all in all we can write down the following we can say in general or rather a lambda let us make it a lambda is equal to 1 by mod a integral minus to plus infinity h gamma g gamma well actually let me let me work this out. So, a t minus a lambda plus c is essentially a t minus gamma now minus gamma means minus a lambda minus b plus b plus c. So, I have a minus gamma here plus a t plus b plus c and a d gamma there let me write this down neatly. So, I am saying essentially it is 1 by mod a integral from minus to plus infinity h gamma g a t plus b plus c minus gamma d gamma and this is obviously the convolution of h and g evaluated at a t plus b plus c. So, we have the answer now h a t plus b convolved with g a t plus c is essentially 1 by mod a h convolved with g evaluated at a t plus b plus c and we can use this in the dilation equation that we had a few minutes ago. Therefore, using this and denoting phi 0 t convolved with phi 0 t as phi 1 t and now I will explain why I am calling it phi 1 t what is phi 0 t convolved with phi 0 t this is phi 0 t when you convolve it with itself it gives you a function that looks like this this is phi 1 t what is 1 about this and 0 about this the degree of the polynomial if you look at it phi 0 t is essentially piecewise polynomial with the polynomial of degree 0 constant phi 1 t is piecewise polynomial with polynomials of degree 1 linear. So, you could similarly conceive piecewise polynomials of higher and higher degree. So, you could have a phi 2 t in fact if you take phi 1 t and convolved with phi 0 t once again you would get a so called phi 2 t a piecewise polynomial with polynomials of degree 2 and if you keep repeating this the degree of the polynomial increases by 1 every time. So, that is the reason why 0 1 2 and so on. Anyway coming back to the dilation equation phi 1 t can easily seem to be half phi 1 t plus 2 phi 1 2 t well phi 1 2 t I am sorry plus 2 times phi 1 2 t minus 1 plus phi 1 2 t minus 2. In fact, what we are saying essentially is that phi 0 t convolved phi 0 2 t convolved with phi 0 2 t minus 1 is the same as phi 0 2 t minus 1 convolved with phi 0 2 t and both of them are equal to half phi 1 2 t minus 1. This is a simple application of the result that we derived a few minutes before and in fact this can be seen graphically. You know if you draw phi 1 t let us draw it large and big it is not very difficult to see that if you were to mark the half points here and draw these triangles this is essentially half phi 1 2 t this is essentially half phi 1 2 t minus 2 and this is essentially phi 1 2 t minus 1 I mean this means this here this means this here and this refers to this here. So, this piecewise linear function plus this piecewise linear function plus this piecewise linear function would obviously give you this solid piecewise linear function it can easily be seen graphically. In fact, linear functions are easy to add you need to add them only at the end points. For example, here the sum is this at this point the sum is essentially this at this point the sum is essentially this. So, you can see that these three dilates and translates of phi 1 t give you back phi 1 t and therefore, phi 1 t obeys a dilation equation in its own right and what are the coefficients of the dilation equation the coefficients in the dilation equation phi 1 t is half phi 1 2 t plus phi 1 2 t minus 1 plus half phi 1 2 t minus 2 are essentially half 1 and half and if you put a filter to represent these coefficients that filter would essentially be the following. This filter is easily seen to be 1 plus z inverse the whole squared, but there is just one short coming here. Unfortunately, phi 1 t is not orthogonal to its integer translates in general. Let me show this to you graphically. If you consider phi 1 t and phi 1 t minus 1 this is phi 1 t and this is phi 1 t minus 1 they are not orthogonal. So, we have a problem although phi 1 t obeys a dyadic dilation equation and in fact, in principle one could also start investigating the basic axioms of a multi resolution analysis. The axiom of orthogonality is not obeyed. So, now we have two options. We can ask whether we can do something similar to what we did in the discretization of scale. If not quite a constant I am talking about the sum of dilated spectra, we said if that sum of dilated spectra is not quite a constant could be allowed to be between two positive constants. So, something similar could possibly be investigated here that is one option. The other option is think out of the box and in fact, we will do that first and come to the JPEG 5-3 filter bank. Why stick to the same analysis and synthesis filters? Can we allow different lengths for the low pass and high pass filter, analysis filter and then therefore, derive the synthesis filters in a slightly different more relaxed way from the analysis filters and establish a perfect reconstruction filter bank and that is what we shall do in the next lecture. Investigate that variant think out of the box, do not insist on orthogonality or do not insist on using the same axioms of an MRA. Build a perfect reconstruction filter bank where you have different length analysis and synthesis filters. I mean different length analysis high pass and low pass filters and therefore, different analysis and synthesis filters and build a different paradigm for multi resolution and this is out of it. So, we shall proceed the next time to derive the JPEG 2005-3 filter bank.