 a very warm welcome to the ninth lecture on the subject of wavelets and multirate digital signal processing. We continue in this lecture to build further on the relationship between the filter bank and the scaling function and wavelet functions. Let me put before you some of the important conclusions that we had drawn towards the end of the previous lecture. We had said that there is a generic dilation equation that relates the filter bank to the scaling function and the filter bank to the wavelet. In fact, if H n is the low pass filter impulse response, we had said that phi t obeys a dilation equation like this. And as far as the wavelet is concerned, we had said that if we take the high pass filter in the filter bank, so if G of n is the high pass filter impulse response, then psi of t is summation n going from minus to plus infinity G of n phi 2 t minus n again phi not not psi. So, this is not surprising what we said was that after all phi t belongs to V 1, psi t also belongs to V 1. So, therefore, both phi t and psi t should be expressible in the basis of V 1 and that is what we have essentially written down. What is noteworthy is that the coefficients of the impulse response of the low pass filter and high pass filter act as the coefficients in the expansion in terms of the basis. Now, in particular for the R MRA, we had noted H n is this sequence. Recall that this is a way of denoting finite length sequences and this means that at n equal to 0, the value of the sequence is 1 and then points after and before take values are shown. So, for example, here if this is n equal to 0, this is going to be n equal to 1 and of course, other points which are not shown are automatically 0. G n is this for the Haar system and in fact, we said that what these equations told us was something much deeper than the containment of phi t and psi t in V 1. In a sense, these equations tell us how to go from the filter bank to the wavelet and from the filter bank to the scaling function. We had just hinted at this in the previous lecture, but now we make that idea very very concrete. So, let us begin by looking at the Fourier domain. As I said last time, we need to take the Fourier transform because that is where we shall see something very interesting. Let us take the first of the two dilation equations. Incidentally, just as you have differential equations, you have difference equations, you have dilation equations here. So, we often encounter differential equations. Example could be y t is a 1 d x t d t let us say plus a 2 x t. It is a differential equation. We have difference equations. For example, y of n is half x of n plus x of n minus 1 is an example of a difference equation which describes a discrete system and now we have a dilation equation. So, this is a new class of equations. It is a new class of equations and this new class has arisen from our discussion of wavelets. In fact, from the relation between wavelets and multirate filter banks. Anyway, with this little aside, let us come back to the issue of relating that filter completely in generative terms to the scaling function and the wavelet. So, let us take this very general dilation equation. Phi of t is summation n going from minus to plus infinity h of n phi of 2 t minus n and we take its Fourier transform on both sides. Indeed, let us denote the Fourier transform of phi t as phi cap omega. Now, remember this is the analog frequency variable or the frequency variable corresponding to the continuous time context. So, I should say analog angular frequency variable to be very precise and we know the relation between phi t and phi cap omega. So, we have phi cap omega is integral from minus to plus infinity phi t e raise to power minus k omega t dt and we operate this on both sides. So, we write down integral from minus to plus infinity summation h n over n times phi 2 t minus n e raise to power minus k omega t dt and integrated all the way from minus to plus infinity. So, we have this integral here. Now, let us you see if this converges which it does because it is the Fourier transform of phi t, we could interchange the order of summation integration. So, we would have this is equal to summation n going from minus to plus infinity h of n integral from minus to plus infinity phi 2 t minus n e raise to power minus k omega t dt. So, we have isolated the part that operates with dt here. Let us evaluate that part separately. Put 2 t minus n equal to lambda where upon we have t is equal to lambda plus n by 2 and of course, one can also write down dt. dt is essentially half d lambda and substituting this we have the integral becomes integral from minus to plus infinity phi of lambda e raise to power minus k j omega lambda plus n by 2 d lambda and a half outside and we can do a little more work on this. So, we keep the terms dependent on lambda inside and we have j lambda rather j capital omega here n by 2 emerging outside. This is j capital omega n by 2 and this is minus infinity to plus infinity phi lambda e raise to power minus j omega by 2 lambda d lambda and this is familiar. This is essentially phi cap evaluated at omega by 2 as one can say the Fourier transform evaluated at the point capital omega by 2. So, now we have a very beautiful relationship. You see what we are saying in effect now is that we can express the Fourier transform phi cap omega in terms of itself which is not surprising because you have a recursive dilation equation on phi t. So, there is a corresponding dilation equation on the Fourier transform. What is that dilation equation? That dilation equation is phi cap omega is summation n going from minus to plus infinity h n times half e raise to the power minus j omega by 2 n times phi cap omega by 2. Now, you know this part of the summation the part of the summation that involves n is familiar to us again indeed we note that summation n going from minus to plus infinity h n e raise to the power minus j omega n would be essentially the DTFT the discrete time Fourier transform of h evaluated at capital omega. So, all that we have done in this expression is that we have replaced capital omega by capital omega by 2 from this point and we will give it we will again use the notation that we have been using. So, we are saying if h of n has the discrete time Fourier transform or DTFT given by capital H of omega. Now, note here I am using the continuous or analog variable that is because I want to retain my discussion in the analog domain or in the continuous time domain. So, I am substituting small omega by capital omega here for the sake of consistency in notation and if h n has the discrete time Fourier transform given by capital H of capital omega. Then what we have here is the following dilation equation the frequency domain dilation equation is phi cap capital omega is half capital H evaluated at omega by 2 times phi cap evaluated at capital omega by 2. You see the beauty is that a dilation equation which involved summation over many terms has now become a dilation equation involving a simple product. The how do we interpret this the Fourier transform of phi t is the same Fourier transform evaluated at omega by 2. So, evaluated omega is equal to evaluated omega by 2 times the DTFT. Now, the beauty is that what we have done here to go from phi cap omega to phi cap omega by 2 can be done to go one step lower. So, the same equation can be rewritten at capital omega replaced by capital omega by 2 and doing that you would have phi cap evaluated at capital omega by 2 is half H evaluated at omega by 4 times phi cap evaluated at omega by 4. So, now we have a recursive process every time you have phi cap omega by 2 you replace it in terms of a product of phi cap omega by 4 and then a DTFT. So, ultimately we have something like this we have phi cap omega is like a product. It is a product m going from 1 to n capital n you like half H omega by 2 raise the power of m this product and then multiplied by phi cap omega by 2 raise the power capital m. So, we have a product of these discrete so called discrete time Fourier transforms here. The only catch is now we need to use the analog frequency variable because we are dealing with analog frequencies here and here. Now, we can take the limit or continue towards n going towards infinity. Now, what is going to happen when you make capital n go towards positive infinity here? Any finite capital omega is going to be taken closer and closer and closer to 0. Again if you wish to be very finicky you should use the opponent proponent model where you say no matter how small I ask this argument to be I can make it small enough and so on. But I think we understand well enough that you can make capital n as large as you desire and you get a larger and larger number of terms in this product and you can take this argument to as small a value as you desire whenever capital omega is finite. For finite capital omega we have the limit as capital n tends to positive infinity of capital omega divided by 2 raise the power of n equal to 0. So, therefore, at least on the finite frequency axis the left hand side is equal to the right hand side well and the right hand side has essentially the Fourier transform of the left hand side at the point 0. So, what do we have here? Let me write that down mathematically phi cap omega therefore, is essentially a product m going from 1 to infinity, positive infinity. Remember the half occurs with each of these terms in the product. Now we have to be careful and say for finite capital omega, but that is not a very serious problem. You see if you look at the Fourier transform of the Haar scaling function for example, let us look at it. Let us look at the Fourier transform of phi t in the Haar context. Essentially it is this and the Fourier transform is easy to calculate. Now we can simplify this using the standard trick of taking out e raise the power minus j omega by 2 term and then we note that we have a sign hidden there and doing away with the j's we will have. Now as you can see this Fourier transform goes towards 0 as capital omega goes towards infinity. So, the Fourier transform vanishes as capital omega goes towards infinity and in fact we can even sketch the magnitude to get a feel of this. The magnitude of phi cap omega looks like this. You see at omega equal to 0 you will notice that one can use L'Hopital's rule and show that this is equal to 1 in magnitude. So, it has a pattern like this of course we know where this comes. This will come at omega by 2 equal to pi or at 2 pi. This would come at 4 pi and so on. This is how the magnitude looks. Now you know this very clearly shows that phi has a concentration around 0 frequency. So, this is interesting. We begin from the low pass filter. We construct a dilation equation, a recursive dilation equation starting with a low pass filter and we get a low pass function. Phi t is essentially a low pass function. A low pass function means it is predominant in the frequency domain around capital omega equal to 0. Low pass function of course is an informal term. You may always argue that after all it does have bands at higher frequencies too, but the point is its prominent bands are around 0 and the farther you go away from 0 the more the spectral magnitude drops off. In that sense it is low pass. In fact just as we tried to build the idea of ideality in a filter bank we also would like to build the idea of ideality in this phi t. The ideal phi t is actually the ideal low pass function. It is a low pass function which is like a brick wall. So, in some sense phi t in the spectral domain or phi cap omega for that matter is moving towards the ideal. You know this is where we are moving and where we are is somewhere here, very far away from it of course. Not surprising after all that is what we saw in the Haru multi resolution analysis too. Ideal here, actual there. Now once again there is the same conflict that drives the engineer the scientist or the mathematician. We know that we want to go towards the brick wall ideal, but we also know the brick wall ideal is unattainable for various reasons. The reasons are similar to what I talked about last time for the unattainability of the idealism in a filter bank. So, we would not need to repeat them once again here. However, what we will now do is to see what is the relationship whether ideal or practical. What is the relationship between phi cap omega and the discrete time Fourier transform of the filter bank low pass filter in terms of construction. So, in other words we have written down a dilation equation in the frequency domain, but we need to translate that dilation equation into a constructive step. How do we construct phi t given the low pass filter impulse response? In a way if we do that we have answered the question how is the design of a filter bank related to the design of a multi resolution analysis. So, let us do that. Towards an objective let me put before you once again that infinite product here. So, you know now you also understand why phi cap 0 need should not be 0, phi cap 0 must be non 0. In fact, phi cap 0 is the very often the maximum value of the magnitude of phi cap omega because of that low pass character. So, we have seen this low pass character in the higher context. Now, we shall assume it to be true of most multi resolution analysis and proceed from there. So, this is just a constant you know a non 0 constant to be varying, but what we need to identify is this. So, we need to focus on this infinite product here phi cap 0 is just a constant. So, let us take just two terms in this product instead of infinite terms. In fact, you know now we need to interpret this continuous analog variable a little more carefully here. When we bring in the idea of a continuous analog frequency variable here, then we need to remember that we are taking the Fourier transform of a continuous function. Now, what is the idea of the discrete time Fourier transform which is of course, of a sequence becoming the Fourier transform of a continuous function? Well that is simple. Suppose you thought of the sequence as a train of impulses located at the integers. So, the sequence h n can be thought of as a train of impulses at the integer locations and the train of impulses therefore, is of course, a continuous function. So, you can take its Fourier transform and use the continuous or analog frequency variable. Now, one must interpret capital H of capital omega in that sense. So, capital H of capital omega is the Fourier transform of this analog function or continuous variable function. Maybe I should say continuous time function to be precise and now what is half h omega by 2 then? For that purpose let us assume that we have a function h of t whose Fourier transform is capital H of omega. Of course, we know capital H omega is integral from minus to plus infinity h t e raise the power minus j omega t dt and if we happen to consider alpha times h omega by alpha with positive alpha. So, what I am saying is consider alpha times h alpha times omega with alpha positive. It is equal to alpha times integral from minus to plus infinity h t e raise the power minus j omega minus j alpha omega t dt and now we have a simple step that we can perform. If we simply put alpha t equal to lambda then we would get you see alpha t equal to lambda alpha is strictly positive. So, when t runs overall from minus to plus infinity lambda also runs from minus to plus infinity. So, therefore, we would have this is equal to alpha times integral minus to plus infinity h lambda by alpha making the substitution e raise the power minus j omega lambda. Now, dt is d lambda by alpha and now if we just cancel the alpha here and the alpha here we get integral from minus to plus infinity h lambda by alpha e raise the power minus j omega lambda d lambda which is essentially the Fourier transform of H of lambda by alpha as the argument. So, we have divided the argument by the positive number alpha. So, what we are saying in effect is if h t has the Fourier transform h of omega then h t by alpha has the Fourier transform alpha times h of alpha omega where alpha is of course, strictly greater than 0 here. Now of course, one can generalize this for alpha real and negative all that one needs to do is to take a modulus outside and no modulus inside, but I leave that as an exercise for you. We do not immediately require it one can easily generalize, but coming back to the point then what h omega essentially means what what h omega by 2 essentially means then with the factor of half outside is a dilated version of this train of impulses. So, we have this train of impulses corresponding to the impulse response h of n which we have called h of the continuous variable t its continuous time Fourier transform or analog Fourier transform so to speak is capital H of capital omega and then half capital H of capital omega by 2 is then the continuous Fourier transform of h of 2 t that is easy to see because you have chosen alpha equal to half there. What you mean by h of 2 t? H of 2 t means you have squeezed h t by a factor of 2 on the time axis on the independent variable. So, you have brought the impulses closer now when you multiply two Fourier transforms the corresponding continuous functions are convolved. So, essentially you may think of h of t here so to speak for the hard case h of t looks like this there is an impulse at 0 and an impulse a continuous impulse remember at 1 these are impulses as understood in continuous time and h of 2 t will look like this you know if I really wish to be finicky I should be putting down the strengths of the impulses carefully 2, but let us not get that finicky. So, this is what h of 2 t will look like there are there are impulses at 0 and half h of 4 t for example, will look like this now this one squeezed again by a factor of 2 there be an impulse at 0 and an impulse at 1 by 4 and so on so forth of course, the rest of it is 0 just 2 impulses. So, what you have now we have a product let us take let us take just 2 terms in that product. So, if we take just the first 2 terms half h capital omega by 2 times half h capital omega by 4 it corresponds to h of 2 t convolved with this is continuous time convolution here. Convolved with h of 4 t now as I said I am being a little careless about constants, but if you really wish to be finicky you can I am more interested in getting a feel of the shape of the convolution I am not so concerned about the precise heights and so on right anyway let me convolve them and show you. So, let us put back h of 2 t and h of 4 t as we had them here. So, we had h of 2 t here essentially 2 impulses located at 0 and half we had h of 4 t with 2 impulses located at 0 and 1 fourth. Now, what will happen when you convolve these you know when you convolve a continuous time function with an impulse a unit impulse it gives you back the same continuous time function. So, as I said if you just ignore the heights are equal here then when you convolve this h of 4 t with this you could treat it as the convolution of this with this impulse plus the convolution of this with just this impulse and you could sum these 2 independent convolutions. When you convolve this with this impulse you simply relocate this at the position 0 and in fact that gives you back h of 4 t when you convolve h of 4 t with this impulse located at half it simply shifts this function to lie at half. So, in effect when you have h of 2 t convolved with h of 4 t you get something like this you get an impulse located at 0 1 at 1 fourth 1 at half and then at 1 at half plus 1 fourth which is 3 by 4. So, you get impulses here. Now, convolve this again to take the next term with h of 8 t as that infinite product asks you to do. So, if you take 3 terms then you would be now convolving this with h of 8 t. How will h of 8 t look h of 8 t looks like this they come even closer together 0 and 1 by 8. I am convolving h 2 t convolved with h 4 t and then the whole convolved with h 8 t what will you have essentially this has to be located here, here, here, here, here, here, here, here, here, here, and here and all these relocated h of 8 t should be added together. Now, when you locate h of 8 t here you will get an impulse at 0 and 1 by 8 here in the middle. When you relocate h of 8 t here you will get an impulse here and at 1 fourth plus 1 eighth that is 2 eighth plus 1 eighth 3 eighth. So, let me straight away now draw this convolution results in impulses at each of these places. Now, you know we seem to be getting where we want to what is happening if you think about it. Each time you bring in one more term you are getting a train of impulses where the train has double the size, but it lies with the same support h of t lay on the support 0 to 1 h of 2 t lies on the support 0 to half. Of course, I would not really say 0 to half you know there is an impulse at 0 and an impulse at half, but then when you go to h 2 t convolved with h 4 t you get an impulse at 0 at 1 fourth at 2 fourth and at 3 fourth. When you go and bring in one more term you get 8 impulses when you bring in one more term next time you are going to get 6 t impulses then 32 impulses and the last impulse comes closer and closer to 1. So, what we have here effectively is h 2 t convolved with h 4 t and so on so forth h 4 t and so on so forth so forth up to h 2 raised to the power of n t is essentially how many impulses you see with when you reached h of 8 t you have 8 impulses. So, when you reach h 2 raised to the power of n t you have 2 raised to the power n impulses located at k divided by 2 raised to the power of n k going from 0 to 2 raised to the power n minus 1. So, you know the last impulse as you can see the last impulse is located at 2 raised to the power of n minus 1 divided by 2 raised to the power of n. So, the last impulse goes closer and closer and closer to 1 the last of these impulses goes closer and closer to 1. So, you know when you have impulses located closer and closer and closer together you are ultimately coming to a continuous function you remember that idea of expressing a continuous function in terms of impulses essentially captures this when you say x of t is a conglomeration of impulses located at every point t with strength equal to the value of x at the point t that is exactly what you are saying when you bring impulses closer and closer and closer together they fuse together to form a continuous function. And it is very easy to see here what continuous function we are moving towards it is flat and indeed it is very clear then that we are moving towards which is essentially 5 t low and behold a very beautiful relationship that we have. We start from the Haar low pass filter we have repeatedly convolved a train of impulses you know first time it is a train of impulses located at 0 and half then at 0 and 1 4th and you have repeatedly convolved these iterated the filter bank repeatedly convolved these trains of impulses and you are moving towards a continuous time function which is indeed 5 t as you can see when you put those impulses closer and closer and closer together. So, now we see the connection between iterating the filter bank and producing phi we now need to complete a little detail how do we get psi. But that is very easy we have already got phi and we know the dilation equation for psi. So, we have phi now how will psi look psi t is essentially summation n going from minus to plus infinity g n phi of 2 t minus n and for the Haar case we know what g n is g n phi is essentially 1 and minus 1. So, we can write down psi t in terms of phi t and construct from there phi 2 t minus phi 2 t minus 1 this is phi 2 t this is minus phi 2 t minus 1 and when we put these two together we get psi t there we are we have completed this i t iteration and building phi t and psi t starting from h n and g n. Now we have a convincing reason to conclude that there is an intimate relationship between the low pass filter and the high pass filter in the two band filter bank and the scaling function and the wavelet function in the multi resolution analysis. In fact, we have constructively established that relation we have shown a procedure by which we can construct phi t and psi t from these impulse responses and therefore, we are now convinced that if we understand how to design two band filter banks and if this iteration is going to converge each time we design a properly designed two band filter bank which allows this iteration we get a new multi resolution analysis. With that background we shall conclude the lecture today and proceed in the next lecture therefore, to explore the two band filter bank more deeply. Thank you.