 A warm welcome to the 24th lecture on the subject of wavelets and multirate digital signal processing. In the previous lecture, we had looked at the admissibility condition for the continuous wavelet transform in depth. The admissibility condition was essentially a condition for reconstructibility from the continuous wavelet transform as we saw. The continuous wavelet transform we noted was extremely redundant to use a continuous scale and continuous translation, meaning essentially a two dimensional representation when what we are dealing with is a one dimensional entity is extremely redundant, very wasteful. And therefore, we began to ask the question of discretization. Can we discretize the scale and can we discretize the translation parameter? We first addressed the question of discretizing the scale and in that context, we had noted what happens when we change the scale. We shall proceed in this lecture today to build on this discretization of scale in greater depth and to consider one particular kind of discretization of scale with which we have been dealing in the first half of the course namely dyadic discretization. Therefore, I intend in the lecture today to talk about logarithmic scale discretization in general and dyadic discretization in particular. With that little background, let us look once again to what the continuous wavelet transform does. We had noted that a continuous wavelet transform essentially operates like a filter both in synthesis and in analysis. So, we had noted that the continuous wavelet transform at scale s is essentially a filtering operation. A filtering operation with a frequency response given by psi cap s omega with some constants of course, at the moment let us simply ignore the constants. So, you see the output independent variable here can be interpreted as the translation. The output independent variable is tau here. Well, then if we take an ideal filter or an ideal wavelet so to speak, we had taken examples of ideal frequency responses that were admissible and those were strictly band limited. They had a psi cap which looked something like this. Of course, I am assuming that the wavelets are real and I am looking only at the positive side of the frequency axis. So, we had noted that the admissibility condition is essentially a statement of band pass character. Therefore, if we consider an ideal band pass function like this and take a dilation of the same, we again get a band pass function. So, for any s greater than 0 as we should have it psi cap s omega. This was of course, the Fourier transform psi cap omega psi cap s omega would look like this and we once again note that there is a contraction or expansion a dilation in general both of the band and the center frequency and that is the reason why we had argued that logarithmic discretization makes sense. Logarithmic discretization means s should be discretized, s should be discretized according to a naught to the power k. So, where k runs over all the integers and we said as long as a naught is positive and greater than 1 and k runs over all the integers, we do not need to take a naught less than 1 also as a possibility because after all when k runs over all the integers, the less than 1 possibility is taken care of. I had given you the example of 2 for example, 2 raise the power of k over all integer k is the same thing as saying half raise to the power of k over all integer k. With that little recapitulation of our discussion, let us go again in depth to analyze what would happen when we discretize the scale. So, in other words when we discretize the scale what we are trying to ask is what do you do to each of these filters and how do you recombine what you have done to each of these filters to get back the original x of t. In other words for each such k, we have a filter and you see how the frequency response of this filter looks. Let us start with the ideal situation and then let us degrade to the practical situation eventually. In the ideal situation the filter the k th filter so to speak would have a frequency response like this. So, it would have a band going from omega 1 divide by a naught to the power k to omega 2 divide by a naught to the power of k where s is of course, a naught to the power k. Now, it is obvious how we can reconstruct this part after all if things are ideal and in fact, if you do pass x t the input through an ideal filter like this to get the continuous wavelet transform here. The reconstruction should also proceed by doing the same thing. So, in fact the k th branch of reconstruction should essentially have the continuous wavelet transform coming in here. The k th branch I mean the continuous wavelet transform being fed to the same ideal filter and if we add this overall k we should get back x t. So, in the ideal situation the answer is very simple. What each branch so to speak each k th branch in some sense of this continuous wavelet transform does is to extract a particular band on the frequency axis and now we can also see very easily interpret very easily what each branch should do. Each branch should have a separate non overlapping band just to take an example. Suppose you had a band pass filter of the following kind. So, if you had psi cap omega looking something like this suppose this were psi cap omega 1 between pi and 2 pi and 0 everywhere else and naturally mirrored on the negative side of omega. Then the k th branch will essentially do the following. The k th branch would cover the band between pi divided by 2 to the power k if a naught were equal to k. So, I must also specify what a naught is k th branch with a naught equal to 2. So, essentially the k th branch would take the frequencies from pi by 2 raise the power of k to 2 pi by 2 raise the power k and we can see that in this particular case for different integers k these bands would be non overlapping. So, it is easy to fix our ideas. In fact, what I have done is to put before you the very idealization of the dyadic multi resolution analysis which we have been looking at all this why. In fact, let us look back at this frequency domain here. You see take for example, k equal to 1 k equal to 1 would take you over the band pi by 2 to pi and if you look at the original band its pi to 2 pi k equal to minus 1 would take you from 2 pi to 4 pi again non overlapping with pi to 2 pi and you can continue in a similar way k equal to 2 k equal to minus 2 and you can proceed k equal to 0 need less to see keeps you in the original band pi to 2 pi. So, now we are beginning to understand what the continuous wavelet transform does in an idealization equations. You see in that case are we really just talking about doing things only in the frequency domain well we must not forget what we are trying to do and which we can never do because of the uncertainty principle is to do what I have just shown you on the frequency axis ideally, but do it with time limited functions. And alas if you look at the inverse Fourier transform of this psi cap homogenous function omega that we just drew here it is far from time limited in fact it decays as slowly as the reciprocal of time terrible anyway. So, this is easy now to interpret and understand and now it is also easy to see that if you take the ideal situation then you could simply put the same filter on the analysis side and the synthesis side. This also makes it clear how we get the notion of filter banks when we discretize. So, you see the discretization of the scale parameter is equivalent to constructing a filter bank recall what a filter bank is. A filter bank is a collection of filters either with a common input or with all the outputs sum together to get a common output and of course more subtly the filters are interrelated. Now all these qualities are satisfied in the k th branches that we are talking about here. In fact in the particular k th branches that we speak of here our analysis filters have the following pattern. The analysis filters are the ones that create the continuous wave led transform strictly speaking we should put a complex conjugate here as was the case with the continuous wave led transform. So, these are the analysis filters and this is the frequency response if you ignore the constants. The synthesis filters again if you ignore the constants have a frequency response that looks like this. This is the k th branch and therefore, if you analyze what you are doing all over the k th branches for k going from minus to plus infinity you are doing the following. On the k th branch if the input signal x of t has the Fourier transform of omega then we are actually constructing x cap omega times psi cap a naught to the power k omega complex conjugate and therefore, this is on the analysis side I mean and on the synthesis side the k th branch multiplies this again by psi cap a naught raise the power of k omega which essentially amounts to multiplying the original Fourier transform by a modulus squared of psi cap a naught raise the power of k omega and on the output side one is going to sum up all these outputs of the branches. So, the final output or overall output is going to be x cap omega multiplied by a summation on all k psi cap a naught raise the power of k omega modulus the whole squared and a minute ago we saw what this was for the ideal situation ideally this should be one one for all omega of course, that is where the challenge is we wanted to be one for all omega with time limited functions and that is the whole challenge. Anyway, how do we relax in the frequency domain if we want to meet this challenge well relaxation means do not quite insist it must be a constant allow it to be between two positive constants. Now, I shall insist on strictly positive constants and we will see why what I am saying is to relax this requirement for design for design ability let this quantity psi cap a naught to the power k omega mod squared summed over all k all integer k be between two constants. So, be greater than or equal to a constant let us say c 1 and less than equal to a constant let us say c 2 and we are told that c 1 is of course, less than equal to c 2 c 2 is less than infinity and c 1 is greater than 0 note this is greater than not greater than or equal to and of course, this is strictly less than not less than that is this cannot go to infinity. So, here there is the possibility of equality here there is not these are strict inequalities that means, c 1 is strictly positive c 2 is strictly finite and of course, it is not difficult to see that this is a non negative quantity. So, it cannot possibly become negative. So, all that we are saying is instead of insisting that this be a strict constant allow it to be between two constants. Now, this is the case then can we make a small change to the synthesis filter and reconstruct that is the first question that we shall answer that is very easy to see. You see in that case let us define another function psi till day from psi now psi till day shall be defined in the frequency domain. So, let psi till day t have the Fourier transform psi till day cap omega. So, we will show that you know even if the wavelet psi t does not quite obey this strict constancy of the sum we can still build a reconstruction approach that is what we are trying to show. So, you see suppose this is the definition of its Fourier transform define psi till day cap omega to be psi cap omega divided by the sum here and please note that since we have this condition here this sum lies between two positive constants and what is important here is the fact that it does not go to 0 the c 1 constant is important here because the c 1 constant this denominator never goes to 0 and therefore, this is valid because of c 1 being greater than 0 otherwise this definition would be meaningless. So, the c 1 is required for this reason to be able to build this psi till day cap. Now, we will show something interesting you see we will show that we could use psi on the analysis side and psi till day on the synthesis side and we will first show that if psi obeys this requirement then psi is admissible. So, we will show that now we use c 1 to define psi till day we will use c 2 to prove admissibility. So, once we have ensured this we also have admissibility let us prove that by admissible I mean we need to consider this integral and we of course, agree that psi t is real. So, let us make the remark here psi is real I mean the it is a real wavelength. So, you see let us break this integral into partial, but finite integrals of the following form summation k running from minus to plus infinity integral from a naught to the power k to a naught to the power k plus 1. Now, let me take an example suppose for example, a naught is equal to 2 then the integral from 0 to infinity is the same as the integral from 2 raise the power k to 2 raise the power k plus 1 with k running over all the integers as k takes negative integer values you are covering the range below 1 as k takes positive integer values we are covering the range from 1 and above 1 that is the interpretation here. So, for example, k equal to 0 would cover the range 1 to a naught in general or 1 to 2 in particular if a naught is equal to 2 and k equal to 2 will cover k equal to 1 will cover the range from 2 to 4 and so on so forth and k equal to minus 1 will cover the range from half to 1 and so on so forth in particular with a naught equal to 2 anyway with that we keep the integrand as it is, but as usual we make a substitution on the integrand. So, put alpha equal to a naught raise the power of k times beta and make a change of variable in the integration now when alpha is equal to a naught raise the power of k times beta this limit simply becomes 1 and this limit becomes a naught. So, we have a naught raise the power of k beta here and of course interestingly d alpha is also a naught raise the power of k d beta and therefore, one can easily see that d alpha by alpha is equal to d beta by beta. So, in place of d alpha by alpha I can simply write down d beta by beta. So, now we have a beautiful summation coming in what it means really is that the integral from 0 to infinity psi cap alpha mod square d alpha by alpha is equal to the integral from 1 to a naught summation if I interchange the order of the summation and the integral which I can do simply because this is a finite integral here summation from k running from minus to plus infinity. In fact, I could interchange this order because I am guaranteed that I have a finite integral here and I am guaranteed that this is converged for all beta it lies between two positive constants. So, it is meaningful to do this and in fact, now we have a condition on this. This is a non-negative integrand here and this is of course, you know integral over a non-negative quantity 1 by beta between 1 and a naught. If I can put an upper bound on this the upper bound on this is known to us it is c 2. So, the summation k running from minus to plus infinity psi cap a naught to the power of k beta mod squared is upper bounded by c 2. And therefore, the so called admissibility integral as we call it we shall now introduce this term the admissibility integral is upper bounded also by integral from 1 to a naught c 2 t beta by b beta and this is a very easy integral to evaluate it is essentially c 2 integrated log natural beta from 1 to a naught and that is c 2 log natural of a naught a neat and elegant result. So, once I have the upper bound on this summation I am guaranteed admissibility. So, we have shown that if psi obeys this requirement and now we will introduce some terminology we shall call this quantity which we are talking about all this while the summation we shall call this quantity a sum of dilated spectra and we shall abbreviated by s d s and just as in the case of the continuous wavelet transform we have primary and secondary arguments for this sum of dilated spectra. The secondary argument is psi because it is for psi that we are constructing a dilated dilated or sum of dilated spectra and of course, the primary arguments are a naught and omega. Well, actually a naught is a grey area you could treated a secondary or primary you know in the continuous wavelet transform for example, we distinguished primary and secondary by calling those arguments primary which did not change in the discussion in a particular context and secondary for those that did or rather the other way around primary for those which did change which were important in a particular context and secondary for those that did not. So, for example, in the continuous wavelet transform the primary arguments where the translation and the scale those were important in a given context they changed in that context and we were looking at the variation of those in a context and the secondary arguments were the essentially the wavelet with which we are constructing the C w t or the continuous wavelet transform and the function on whom the continuous wavelet transform is being constructed. So, here too we shall talk about s d s with secondary arguments of psi and a naught and primary argument omega but of course, we could shift this and what we are essentially saying is s t s s t s psi a naught as a function of omega needs to be upper bounded and lower bounded and the upper bound guarantees admissibility. Now, we will soon see in a minute that this C 1 guarantees reconstruction and in fact we are going in that direction. We have already constructed a psi tilde where psi tilde cap omega is essentially now with this language psi cap of omega divided by s t s psi a naught omega and it is very easy to see that if s t s obeys the upper lower bound condition psi tilde is also going to be admissible. So, we show that psi tilde is admissible and to prove that all that we need to do is to try and look for bounds on the sum of dilated spectra of psi tilde. So, we need to consider for this purpose the sum of dilated spectra psi tilde a naught as a function of omega and this is not at all difficult to compute. Indeed, let us in particular construct psi tilde cap a naught raise the power k omega which is the typical term required or rather the modulus of this squared and the modulus of this squared is very easily seen to be the modulus of psi cap a naught raise the power of k omega the whole squared divided by that is the interesting thing. Let me expand the denominator. I will first write a summation on L to distinguish it from this k. Summation on L going from minus to plus infinity psi cap a naught raise the power of L. Now, in place of omega I need to write a naught raise the power of k omega. So, I will write a naught raise the power of k omega there and now this is easy to evaluate. In fact, you can see that this becomes a naught raise the power L plus k and please remember the summation is on L and k is fixed. So, for a fixed k when L runs over all the integers as it does here L plus k also runs over all the integers. So, what we have in the denominator is once again the sum of dilated spectra of psi a naught evaluated at omega. So, that makes my life very easy. What I am saying in effect is psi tilde cap a naught raise the power of k omega mod squared is simply psi cap a naught the power of k omega mod squared divided once again by s d s psi a naught evaluated at omega. It makes my summation very easy to do. So, what I am saying in effect is that the sum of dilated spectra of psi tilde with parameter a naught evaluated at omega is in other words summation only on the numerator. You see the denominator is independent of k. In fact, the little correction which we need to make here you see here I need to put a square. So, although I had discussed this in the context when I took the modulus squared I should also put a square here. I do not need to put a modulus because this is anyway non negative. So, with that little correction what I have in the denominator in fact, here also I need to make the correction. So, I have s d s psi a naught omega squared here little correction is required. So, here we have psi a naught omega squared and now this is familiar this is essentially s d s psi a naught once again. So, what I have in effect is that s d s psi tilde a naught evaluated at omega is just s d s psi a naught evaluated at omega divided by s d s squared and this is where the lower bound becomes important. Since this is lower bounded by c 1 never goes to 0 and it is also upper bounded by c 2. So, never goes to infinity it is valid to cancel. Cancellation from the numerator and denominator is valid because of the bounds and on cancellation we shall get s d s the sum of dilated spectra of psi tilde evaluated at omega with parameter a naught is just the reciprocal simple. In fact, now I can go a step further I can use the bounds on the sum of dilated spectra for psi to come up with bounds on the sum of dilated spectra for psi tilde let us do that. So, let us come back to this expression you see all that we need to do is to note that if s t s psi a naught omega is bounded between c 2 and c 1 this one greater than 0 and this one less than infinity then I can take the reciprocal on all sides because these are all non negative quantities and I can get infinity is greater than 1 by c 1 is greater than or equal to 1 by s t s psi a naught evaluated at omega is greater than equal to 1 by c 2 which is in turn greater than 0 and this of course is the sum of dilated spectra that we are looking for here this is nothing but s t s psi tilde a naught evaluated at omega. So, we have a bound the bounds on the sum of dilated spectra of psi give you the bounds on the sum of dilated spectra of psi tilde and the same arguments that we used to show that psi is admissible can be used to show that psi tilde is also admissible. So, psi tilde is also an admissible wavelet and therefore, I could now construct an asymmetric analysis and synthesis paradigm on the analysis side I would use the wavelet psi on the synthesis side I would use the wavelet psi tilde. So, all that I am saying is use psi tilde on the synthesis side that means the kth branch is as follows it has a filter with frequency response given by psi tilde cap a naught raised to power of k omega and now we can see what happens to x when it goes through the analysis and synthesis side with this analysis and synthesis together would produce x cap omega times psi cap a naught raised to the power of k omega complex conjugate multiplied by psi tilde cap a naught raised to the power of k omega and this is easily seen to be of course, this is summed over all k this is easily seen to be x cap omega times sum for k going from minus to plus infinity psi cap a naught raised to power of k omega complex conjugate psi cap a naught raised to power of k omega divided by the sum of dirated spectra psi a naught evaluated at omega and once again you can see what all this is leading to this is independent of k. This simply gives you x cap omega into sum of dirated spectra psi a naught omega divided by the sum of dirated spectra psi a naught omega and once again because of c 1 and c 2 we can cancel for all omega and that gives us just x cap omega. So, we reconstruct now we have also brought in a new notion here if we have made this relaxation namely that the sum of dirated spectra can lie between two positive constants we also need to generalize our notion of analysis and synthesis a little bit. We need to generalize it by allowing a different wavelet on the analysis side and on the synthesis side. You have psi on the analysis side psi tilde on the synthesis side. So, we are slowly leading to a different paradigm in the context of filter banks all this while the wavelet on the analysis and the synthesis side has been the same. Now, we are allowing them to be different if we want to relax the context of design anyway that remark apart. Let us come back to one specific case the specific case of a naught equal to 2 in the specific case of a naught equal to 2 which is also called the dyadic case dyadic refers to this a naught equal to 2 what we are asking for is that the sum of dirated spectra of psi 2 omega must be upper and lower bounded now in particular if one considers the higher wavelet one can show that this is true I do not wish to go through the details, but rather to leave it to you as an exercise. So, exercise for the higher wavelet for example, show that the sum of dilated spectra as we have constructed lies between two positive bounds. Now, you know to give you a hint it is clear that the upper bound essentially gives you admissibility and it is not very difficult to show the higher wavelet is admissible. So, first if you can simply show the higher wavelet is admissible then this is kind of done the lower bound is a little more tricky and that I leave as an exercise for the class to do and of course, this also holds for the Gaussian wavelet for example. So, I also put down the following exercise for the Gaussian wavelet or the derivative of Gaussian one can once again show a very wide range of a naught is possible. So, one can show that the sum of dilated spectra in the context of the derivative of Gaussian kind of wavelets is always between two bounds for a very wide range of a naught and I encourage you as an exercise to find out what that range of a naught is. Now, one more remark in case the sum of dilated spectra is 1 is constant for all omega. So, in other words c 1 becomes equal to c 2 then we have the situation where psi and psi till they are the same and that is essentially the situation of orthogonal filter banks the kind of filter banks that we have been talking about all this while. So, in other words the specific situation where analysis and reconstruction proceeds from the same wavelet is the case where the sum of dilated spectra well it you know please remember here we are talking about a continuous translation parameter. So, we have not yet discretized the translation parameter. So, it does not follow that the higher wavelet has the sum of dilated spectra equal to a constant because we have discretized the translation parameter there, but if you are not to discretize it then those wavelets where c 1 is equal to c 2 give you an orthogonal filter bank that means the wavelet on the analysis and the synthesis side is the same and in particular when a naught equal to 2 it gives you a dyadic and resist synthesis system. With this then we conclude the lecture today and we shall build further on the dyadic case in the next lecture. Thank you.