 A warm welcome to the thirteenth lecture on the subject of wavelets and multivariate digital signal processing. We continue in this lecture to build upon the particular class of filter banks which we had introduced in the previous lecture namely the conjugate quadrature filter bank. A number of issues related to that filter bank were left unanswered in the previous lecture. To some extent our introduction of the filter bank seemed ad hoc at points. What I mean by that is we had suddenly made little twists in the nature of the filters where a proper justification had not been given simply because there was a bit of a chicken and egg problem. The justification was best seen after we went through the discussion and that is what I had promised that after we complete an understanding of this filter bank many things will be a little more clear. So, let us embark then upon that filter bank once again let us look at that conjugate quadrature structure once again first in Toto and then in specifics. So, in today's theme we shall look at conjugate quadrature filters in depth and we shall again consider one specific class of those conjugate quadrature filter banks namely the family of filter banks and family of multi resolution analysis that emerge from Dobash filters. Incidentally, as I mentioned Dobash or you know sometimes it is pronounced as Dobashis has been a mathematician, scientist, engineer whatever you want to call her of repute. Her important contribution in this field has been to propose a family of compactly supported wavelets which also have some other interesting properties. It turns out that the hard wavelet is the baby of the Dobash family the simplest of the Dobash wavelets and there are further and further ones of which we shall give an introduction today. In fact, the central idea in the Dobash family is to build upon what we had briefly mentioned in the previous lecture namely the idea of keeping and annihilating polynomials of higher and higher degree on one of the two branches of a filter bank. Anyway, we shall look at specifics as we go along, but this is to put the lecture in perspective. So we shall talk today about the conjugate quadrature filter bank and we shall look specifically at the Dobash family of MRA. Now you see the conjugate quadrature filter structure as we understood it at the following relationships between the filters. We had the analysis high pass filter was related to the analysis low pass filter by the following relationship and we had promised that we shall understand this a little better today. Of course, the synthesis filters were related very easily to the analysis filters. So you had g 0 z being h 1 of minus z and g 1 z being minus h 0 of minus z. This of course, was essentially Elias cancellation for you these two conditions. But now let us focus on this relationship of h 1 to h 0. So first let us justify why it is a high pass filter. So let us consider this expression z raised to the power minus d h 0 minus z inverse and let us put z equal to e raised to the power j omega as we do to obtain a frequency response where upon we will have e raised to the power minus j omega d h 0 minus e raised to the power minus j omega. Now if we take the magnitude of this as is normally what we are interested in we have the magnitude of e raised to the power minus j omega d h 0 e raised to the power minus j omega is the same as the magnitude of h 0 e raised to the power minus j omega minus I am sorry minus of this. That is because the magnitude of this is 1 and now let us look at this quantity the magnitude of h 0 minus e raised to the power minus j omega. You see if h 0 is a filter with real coefficients. So if h 0 z corresponds to a filter with a real impulse response and that is the class in which we are most interested. In that case h 0 e raised to the power minus j omega is going to be h 0 e raised to the power j omega complex conjugated that follows in a straight forward way from some basic properties of the discrete time Fourier transform. What we are saying essentially is that the magnitude response of a filter with real impulse response is symmetric in omega and the phase is anti symmetric. If we now replace e raised to the power minus j omega by minus e raised to the power minus j omega what are we really doing. So h 0 minus e raised to the power minus j omega is essentially h 0 e raised to the power minus j omega plus minus pi we have done this before we have noted that minus 1 is essentially e raised to the power plus minus j pi and therefore, what we have done here is essentially to shift this by pi either forward or backward it does not make any difference because there is a periodicity with a period of 2 pi anyway. What we do know is that a low pass filter I think we have seen this quite frequently now a low pass filter when shifted by pi on the frequency axis becomes a high pass filter of course, a low pass filter aspiring to be a low pass filter with a cut off of pi by 2 it becomes an aspirant for a high pass filter with a cut off of pi by 2 again and similarly when a high pass filter is shifted by pi on the omega axis it becomes a low pass aspirant with a cut off of pi by 2 we have seen this pretty much before anyway recognizing this then we have an interpretation for what we just did. So, we said this minus essentially shifts by pi and therefore, h 0 e raised to the power minus j omega without the minus sign would have been a low pass filter as it is because of this conjugate symmetry that we have here and now with the introduction of a minus sign it becomes a high pass filter. So, we have a convincing argument now that h 1 z the way we have constructed it. So, we have convinced ourselves we have shown h 1 z in the way that we have constructed it namely z raised to the power minus d h 0 minus z inverse is indeed high pass or a high pass aspirant aspires to be an ideal high pass filter with cut off pi by 2 provided of course, h 0 z is an aspirant to be a low pass filter with cut off pi by 2. So, now things have fallen into place the only issue is why have we taken this peculiar expression not so peculiar really now we do not see it is so peculiar, but why the z inverse and so on. So, we will understand that in a minute you see I will just give you a trailer for the reason the trailer is that this automatically brings a condition on the magnitude we will see that shortly anyway. Now, let us put down the Elias cancellation condition is anyway put down we need to put down the perfect reconstruction condition. So, let us put down the perfect reconstruction condition we did that yesterday, but we will do it little more carefully the perfect reconstruction condition essentially says that you would have g 0 z h 0 z plus g 1 z h 1 z must be some constant we called it c 0 times z raised to the power minus d and what are g 0 g 1 h 0 h 1 here. Now, we have agreed that g 0 z is essentially h 1 of minus z. So, we had h 1 minus z h 0 z plus now g 1 z we had agreed to make minus h 0 minus z and h 1 z of course, we have agreed to make it z raised to the power minus d and so on, but let me write h 1 z for the moment and we want this whole thing to be c 0 z raised to the power minus d. Now, we will substitute h 1 z in this equation we have minus 1 to the power d z raised to the power minus d h 0 z inverse times h 0 z minus. Now, again you have h 0 minus z here and h 1 z becomes z raised to the power minus d h 0 minus z inverse this you desire should be c 0 z raised to the power minus d. Now, you know this z raised to the power minus d that we have here. In fact, we should not quite have written it like this though what we have written now happens to be correct. We should have started by giving a different value for the delay here and the delay on this side, but now again through serendipity or through convenience we can actually make them the same. The purpose of putting this z raised to the power minus d here was actually to take care of this term here. So, it is not coincidental that we have written the same d on both sides. So, that should not have been done initially we are doing it right away to emphasize that this z raised to the power minus d term that we introduced in h 1 was meant to take care of this. So, what we are saying in effect is that we want the rest of it to match as well. So, what we desire for perfect reconstruction is essentially this minus 1 raised to the d h 0 z h 0 z inverse minus h 0 minus z h 0 minus z inverse is a constant. Now, again we have the freedom to choose the value of capital d here. Again the main issue is whether capital d is odd or even. If capital d is odd then we have a minus in both places for both the terms. If it is even then this is a plus and this is a minus. Let us choose to make capital d odd and in fact again there is a reason for that it is not arbitrary. You know we have just looked at the Haar filter bank where we have a filter of even lengths a length 2 filter actually all of them you know 1 plus z inverse 1 minus z inverse on both sides are of length 2. When we replace z by z inverse. So, let us take the Haar case once again you had h 0 z of the form 1 plus z inverse whatever forget the by 2 here in the Haar case h 0 z inverse would have been 1 plus z and the z to the power minus d h 0 z inverse or if you choose you know you can write minus z inverse as we do and this would then become minus here z raised to power minus d h 0 minus z inverse is actually intended to make this causal this filter is non-causal. So, you need to introduce a z raised to power minus 1 here to make this causal and therefore d becomes 1 in this case. So, you see the role of d I had hinted at this yesterday you see we said that the reason why we cannot avoid a delay is because you want the filters to be causal now you see what we mean this z raised to power minus d term has been put there to retain causality and you just put as much of a d as is needed to allow for causality and. So, here the d required is 1 now in the Dovash family we keep augmenting the filter length by 2 in every rung of the family ladder. So, when we go from the baby of the family namely the Haar MRA to the next member of the family we augment the length by 2. So, we have a length of 4 when we go to a length 6 it gives us the third member and so on length 8 the fourth member and so on so forth. So, successive even lengths of filters give us successive members of the family in the Dovash family now what we are going to do is slowly move towards building the second member of the Dovash family and therefore the next case would be capital D equal to 3. So, you would have a length of 4 and you would have a maximum power of z equal to 3 z cubed when you write h 0 minus z inverse that is the role of z raised to power minus d here. Therefore, it is justified for us to begin by assuming that d is odd. So, let me put that down once again for you in this relationship that we have here we shall now assume d to be odd with d odd we essentially have for perfect reconstruction h 0 z h 0 z inverse plus h 0 minus z h 0 minus z inverse is a constant let me explain you see when d is odd then both of these are minus sign. So, you can take away the minus sign from the left hand sign and put it on the right and this is anyway a constant. So, negative of a constant is also a constant. So, there we are now what does this mean we need to reflect on it little we will first reflect on it in the frequency domain. So, when we put z equal to e raised to power j omega what do we have here h 0 z or rather h 0 e raised to power j omega times h 0 e raised to power minus j omega plus h 0 minus e raised to power j omega plus h 0 minus e raised to power j omega h 0 minus e raised to power minus j omega is a constant. Now, once again we shall remove the minus sign here and shift omega by pi and we shall also note that if you have a filter with a real impulse response then h 0 e raised to power minus j omega is essentially the complex conjugate of h 0 e raised to power j omega. The same holds here when you have omega replaced by minus omega here you again get a complex conjugate of this. So, all in all for real filters we have h 0 e raised to power j omega h 0 e raised to power j omega complex conjugate plus h 0 e raised to power j omega plus pi plus minus if you please h 0 e raised to power j omega plus minus pi complex conjugate is a constant. Now, we have a very beautiful conclusion here you see this is the magnitude squared and this is again a magnitude squared. So, there we are what we are saying in effect is mod h 0 e raised to power j omega squared plus mod h 0 e raised to power j omega plus minus pi the whole squared is a constant. Now, this is very interesting this is exactly one of the properties that we had introduced in the context of the Haar system namely the property of what is called power complementarity. Here it is clear now that by this construction we have achieved power complementarity in the high pass and low pass filters of the analysis side and in fact it is a simple consequence that if we look at the synthesis side they are also power complementary. In fact I leave it to you as an exercise by using the relation between g 0 g 1 and h 0 to show that the synthesis side is also power complementary. So, what do we have here it is very interesting the analysis filters are power complementary and so to the synthesis filters. So, as I said exercise show this we have already proved it more or less it is just a little bit of as they say dotting your eyes and crossing your t's you need to write down need proof, but I think that is a good thing to do we must leave a couple of exercises for the class to do and this is a very simple exercise with which we begin. Use the discussion that we just had over the last couple of minutes to work out the details to show that the analysis filters and the synthesis filters are both a cop a power complementary pair. Anyway this is the motivation for that so called quote unquote peculiar choice of h 1 now we see things falling in place. The z inverse was required to bring this complex conjugation replace omega by minus omega and of course as you see for a real impulse response it had no effect on the magnitude, but we could remove the phase. So, it is a strategic choice of analysis high pass you could have chosen h 0 minus z or something like that, but you chose h 0 minus z inverse because you wanted that complex conjugation and then you put a z raise the power minus d because you wanted to make it causal. So, a z raise the power minus d is to introduce causality the z replaced by z inverse is to introduce to bring in this complex conjugation to bring in power complementarity and finally, the minus I mean minus z inverse instead of just z inverse is to convert the low pass to a high pass. So, now it all falls in place and we have justified our choice and now we also know what we demand of h 0 z so that we get perfect reconstruction. Let us look at that condition once again that condition tells us and let me write it slightly differently. That condition tells us for perfect reconstruction some interesting intermediate filter which we shall define by kappa 0 z. So, let us define kappa 0 z as h 0 z h 0 z inverse what we are saying is that for perfect reconstruction we require kappa 0 z plus kappa 0 minus z to be a constant. Now, things are beginning to make even more self. If we know the sequence that gives us h 0 z what is the sequence that gives us h 0 z inverse let us reflect a minute on this. So, what I am trying to say is we have agreed that we are going to choose even length h 0 z something like an impulse response of the following form h 0 h 1 and so on h 0 lies at 0 up to h 0 z inverse. So, h 0 h d remember d was odd and therefore, h 0 z inverse would then correspond to the following quite clear when you replace z by z inverse you are essentially reflecting the sequence about the point n equal to 0 simple. Now, h 0 z times h 0 z inverse corresponds to their convolution you know when you multiply two z transforms the corresponding sequences are convolved and therefore, we have this convolved with this maybe I should put parenthesis here and indicate the 0 clearly there. Now, how do you convolve? Well these are of equal length. So, I could choose either of them as the static one and the other one as the moving one. So, just for convenience what I will do is the sequence which we started with the one corresponding to h 0 we shall keep as the static sequence and the one corresponding to h 0 z inverse we shall make it move. Now, what we are saying essentially is keep this static. So, you have and make this move. So, when you make the other one move you are doing two things you are bringing you see you want to if essentially you have sequence one let us say sequence let us call that sequence g n just for the time being the sequence g n is this or g k if you like in which case the sequence g n is the sequence n minus k this is of course, a function of k. So, k equal to 0 its h 0 and so on. So, g of n minus k is going to look like this the 0 would go to n and whatever comes before 0 would go after n there. So, you have h 1 and so on up to h d. So, this reaches the point n plus t here this is the sequence g n minus k and this you may of course, call the sequence h 0 of k if you like. So, you are trying to convolve this sequence with essentially with this sequence, but in that convolution you are going to move around this at different locations here. Now, visualize let me put that down clearly once again for you. We are saying we have this so called static sequence and this is going to move around n is moving. So, you can visualize the situation for different values of n this lies at different locations with respect to the static sequence. For example, when n is equal to 0 the samples actually coincide. When n is equal to 1 then h 0 clashes with h 1 and of course, h d has gone out of range. So, it has gone to a 0 sample here. When n is equal to minus 1 you are here and then of course, h 1 clashes with h 0 h d with h of d minus 1 here and so on so forth. So, you see what we have is actually the dot product of the sequence and its own shifted versions. This is very interesting. What we are saying is that the samples of kappa 0 are actually dot products of the original filter impulse response shifted by different amounts of shift. Let us write that down. That is a very important conclusion kappa 0 kappa 0 z which is h 0 z times h 0 z inverse corresponds to a sequence whose nth sample is as follows the dot product of the impulse response corresponding to h 0 and the same shifted by m samples. If you want to be very specific you should say m samples forward, but that does not really matter. So, if you want to write it down in the notation of dot products what we are saying is that this denotes the dot product of sequences a and b. So, you know a with an argument integer argument b with an integer argument this is the dot product of a and b and we are saying the m sample of the filter kappa 0 is essentially the dot product h 0 and h 0 shifted by m plus or minus is not really an issue. If you like you can make this minus. There is a symmetry you know you can visualize that if you shift backward by 2 or forward by 2 it is the same let us verify that for a length 4 for example you will see what I mean. So, if you had a length 4 for example you would have h 0 h 1 h 2 h 3 and if you took this and the same thing shifted by 2 you are talking about this dot product the rest of it is 0 of course. So, here again you get 0s and you do not need to write that. So, the dot product is essentially h 0 h 2 plus h 1 h 3 now if you were to shift it backwards. So, you had h 0 h 1 h 2 h 3 there and you shifted it backwards and of course, this is all 0. So, again the dot product would be h 0 h 2 plus h 1 h 3. So, as you can see shifting backward or forward by m is not an issue. However, what we are saying here is something very interesting we are saying that with this understanding of the samples corresponding to kappa 0 z kappa 0 z plus kappa 0 minus z is a constant and if we take the inverse z transform now and if you only care to multiply by half on both sides this is also a constant obviously and this is something very familiar to us. We have encountered this when we did down sampling. So, in fact if the original sequence corresponding to kappa 0 z. So, you know let kappa 0 z correspond to the sequence let us write small k 0 n then what we are saying is that kappa 0 when this sequence is modulated by a sequence which is 1 at the even locations and 0 at the odd locations. So, it is something interesting we are doing we are modulating this kappa 0 n by a sequence which is 1 at the even locations and 0 at the odd locations. This gives us a sequence corresponding to the inverse z transform of a constant which is essentially an impulse. Now, you know this modulation is what we derived when we talked about the z transform across a down sampler. So, remember when we go across a down sampler by a factor of 2 it is like first modulating by a sequence which is 1 at the even locations and 0 at the odd locations. In general when you go across a down sampler by a factor of capital M it is like modulating with a sequence which is 1 at all multiples of capital M and 0 elsewhere followed by an inverse up sampling operation. So, remember a down sampling by 2 was modulation by a periodic sequence with period 2 which was 1 at locations equal to multiples of 2 and 0 else followed by an inverse up sampler by a factor of 2. Inverse up sampler means a compressor throw away the zeros down sampling by a factor of M was essentially multiplication by a periodic sequence period capital M 1 at all multiples of M 0 elsewhere followed by an inverse up sampler by a factor of capital M which means throw away the zeros and compress. So, we have you see that throwing away the zeros was what made z replaced by z raise the power half. So, here in this expression kappa 0 z plus kappa 0 minus z we are not writing z raise the power of half. So, we do not do that inverse up sampling operation, but the rest of it is there and that is the justification for this step here modulation with this periodic sequence and now this is equal to a constant which means if we take the inverse z transform here we are saying this is essentially the impulse which means this has a non zero value at 0, but 0 everywhere else. So, let us write that down kappa 0 N when modulated with this periodic sequence with period 2 with the ones at multiples of 2 and 0 elsewhere results in a sequence which is non zero only at z equal to at N equal to 0 that is what we are saying and obviously at the odd locations anyway it is 0. So, there is nothing very surprising here it is at the even locations that we have a surprising result there. So, the surprise is at the even locations of course m not equal to 0. So, what we are saying is that if I take the impulse response of the low pass filter on the analysis side shifted by any even number of samples 2 4 minus 2 minus 4 6 minus 6 and so on and take the dot product of that shifted impulse response with the original impulse response that dot product is 0. For those of us who are familiar with the idea of auto correlation what we are saying is that the auto correlation of the impulse response of the low pass filter is 0 at the even locations other than 0. Let us use this to build the first of the family of the Dabash filters you see well I should say first non trivial. So, it is second in that sense the first non baby member Dabash filter with length 4 is going to look something like this it is going to have an impulse response H 0 H 1 H 2 H 3 and recall what we did yesterday we said that in this filter we would need to bring in one more factor of the form 1 minus z inverse in the high pass filter. So, her had 1 1 minus z inverse in the H P M. So, this length 4 filter would have 2 factors 2 1 minus z inverse in the high pass filter and that means you see the high pass filter was obtained by replacing z by z inverse and then by minus z as well. So, if the z inverse part gets taken care of by the delay z raise the power minus d, but the z replaced by minus z needs to be undone to go to the low pass and therefore, the low pass filter would have a factor 1 plus z inverse square. Now, when you say it has a factor 1 plus z inverse the whole square you already constrained 2 of the 3 zeros that it has free to be chosen. What I mean is if you looked at H 0 z it would have been H 0 plus H 1 z inverse plus H 2 z raise the minus 2 plus H 3 z raise the power minus 3. So, there are 3 zeros to be chosen out of them we have already chosen 2. So, we have only 1 free let that free 1 be at b 0. So, in all it is very simple we can take H 0 z to be of the form 1 plus z inverse the whole square times 1 plus b 0 z inverse. What do we have then let us expand this we have H 0 z is essentially 1 plus 2 z inverse plus z raise the power minus 2 times 1 plus b 0 z inverse and we can expand this further that product will be 1 plus 2 z inverse plus z raise the power minus 2 plus b 0 z inverse times this. So, b 0 z inverse plus 2 b 0 z raise the power minus 2 plus b 0 z raise the power minus 3. So, in a sense we have the following impulse response for the filter 1 2 plus b 0 z inverse plus 2 b 0 z inverse 1 plus 2 b 0 and b 0 here this is the impulse response. Now, we have set up the low pass filter for the second member in the dobash family where do we go from here. We shall use the constraint that we just derived namely that the dot product of this impulse response with its shifts by even shifts must be 0 and we shall see the constraints that emerge on the free parameters. In the next lecture therefore, we shall constrain the value of b 0 make a choice for b 0 and derive precisely the impulse response of the dobash second member and thereby also establish a general procedure for building up the dobash family low pass filters. Concurrently we shall explain how this family evolves and recall again the significance of going from one member to the other with that then we shall conclude the lecture today. Thank you.