Wavelet Transform (779450), страница 4
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One chooses thecoefficients of a PR two-channel filter bank in such a way that the waveletsand scaling functions associated with these filters have the desired properties.Scaling Function. The starting point for constructing scaling functions isthe first part of the two-scale relation (8.102):+(t)=cgo(n)d5- n).(8.119)nIn the following the Fourier transform of equation (8.119) is required, which,using1 W -jwn.(8.120)+ ( 2 t - n ) t)- @(-) e 2 ,22is(8.121)(8.122)nequation (8.121) is@(W)1‘ W= - Go(e3T)WJz(8.123)Since the scaling function +(t)is supposed to be a lowpass impulse response,we may introduce the normalization(8.124)If wenow apply (8.119) and (8.123) K times, we obtain(8.125)238TransformChapter 8.
WaveletNowwe let K + 00. If the product in (8.125) converges for Kcontinuous function, it converges toc@(W)m*l- Go(ejw/2k),=k=l+ CO to a(8.126)Jzbecause we have specified @(O) = 1. Thus (8.119) allows us to determine thescaling function recursively. When starting with1 for 0 5 t < 1,0 otherwise,(8.127)we obtain the piecewise constant functions xi(t) by means of the recursion%+l( t )= -Jzcg o ( n ) xi (2t - n ) ,(8.128)nwhich approaches the scaling function for i+ CO.Figure 8.12 illustrates therecursive calculation of the scaling function $(t).However, the convergence of the productdoes not guaranteethat the obtainedscaling function is smooth. Figures 8.13 and 8.14 show examples leading tosmooth and fractal scaling functions, respectively.Wavelet.
If the scaling function $(t) is known, $(t) can be calculated byusing the second part of the two-scale relation (8.102):,$(t)= &1(n)Jz $(2t - n).(8.129)nIt is obvious that a smooth $(t) results in a smooth ,$(t),regardless of thecoefficients g 1 ( n ) , so that all concerns regarding smoothnessare related to thelowpass go (n).Summary of Construction Formulae. According to (8.126), the synthesisscaling function is related to the synthesis lowpass as@(W)=n JzO01- Go(ejw/2k)(8.130)k=lFor the synthesis wavelet we get from (8.129) and (8.130)!@(W)1= - G1(ejwI2)Jzk=2lW- Go(eJz) .(8.131)8.6. Waveletsfrom Filter Banks2391Figure 8.12.
Recursive calculation of the scaling function q5(t);the first two stepsof the recursion are shown (coefficients: {gO(n)} =1, f}).${a,The analysis scaling function $(t)and the wavelet G(t)are related to thetime-reversed and complex conjugated analysis filters h:(-n) and h; (-n) inthe same way as $(t)and ~ ( tare) related t o go(.) and g1(n). Thus, theymaybe given in the frequency domain as(8.132)andChapter 8.
WaveletTransform240024014IE0l0b2400Figure 8.13. Recursivecalculation{ g o ( n ) } = ${l 3 3 l}).2t-2O202of the scalingfunctiontt-4-4(t) (coefficientsM0 PL 2E4I02t-4Non-Linear Phase Property of Orthonormal Wavelets. In Chapter 6we haveshown that paraunitary two-channel filter banks have non-linearphase filters in general. This property is transferred directly to the scalingfunctions and wavelets constructedwiththesefilters.Thus,orthonormalwaveletshave non-linear phase in general.Exceptions aretheHaarandShannon wavelets.2418.6.
Wavelets from Filter Banks8.6.2Requirements to be Metby the CoefficientsWe havealready shown that to constructbiorthogonalandorthonormalscaling functions and wavelets the coefficients of PR two-channel filter banksare required. But, in order to satisfy (8.124), the coefficients must be scaledappropriately. The correct scaling for the lowpass can be foundby integrating(8.119):00$(t)dt=Lc g o ( n ) / O O$(2t - n) d(2t).(8.134)f i nThis yields-00cgo(n) =h.(8.135)nBy integrating equation (8.129) we obtainand with (8.124) and J $ ( t ) dt = 0 we conclude(8.137)nThis means that the highpass filters in the two-channel filter bank must havezero mean in order to allow the construction of wavelets.8.6.3Partition of UnityIn order to enforce a lowpass characteristic of Go(z), it is useful to requireGo(-l) = 0t) C ( - l ) n g o ( n )= 0.(8.138)nAs will be shown in the following, (8.135), (8.135),and (8.138) result in@(27rk) = { l.:;(8.139)In the time domain, this property of the scaling function, which is known asthe partition of unity, is writtencMn=-m$(t - n) = 1.(8.140)242TransformChapter 8.
Wavelet(8.141)={ f ( ~ k )k even,k odd.For Ic = 0 , 1 , 2 , 3 , 4 ,. . . we obtaink = 0 : @(O)= @(O)k = 1 : @(27r) = O . @ ( 7 r )k = 2 : @(47r) = 1.@(27r)= 1,=o,=o,k=3:@ ( 6 ~ ) = 0 . @ ( 3 ~ )= 0,k=4:@(87~) = 1 .@(4~) = 0,(8.142)We may proceed in a similar way for the negative indices, and it turns outthat (8.139) holds. 08.6.4The Norm of Constructed Scaling Functions andWaveletsWhen the coefficients gO(n) belong to aparaunitary filter bank, (8.124)directly leads to(8.143) 11411 = 1.We realize this by forming the inner product of (8.140) with &,(t) and bymaking use of orthogonality:(8.144)@(0)=1(400:400)Forming the inner product ($oo, $oo) by using (8.101) yields243from8.6.
Waveletswhich shows that11+, 1 1(8.146)=1for the norm of the wavelet $(t).Assuming Q(0) = a leads to11q511= 11+11= a.In the biorthogonal case the relationship between the norm of the coefficients and the norm of the scaling function is much more complicated.8.6.5MomentsMultiresolution signal decompositions are often carried out in order t o compress signals, so that the compaction properties of such decompositions areof crucial importance. Most signals to be compressed are of a lowpass natureand can be well approximated locally by low-order polynomials. Therefore, itis useful t o seek wavelets with good approximation properties for low-orderpolynomials.
As we shall see, the approximation propertiesof a multiresolutiondecomposition are intimatelyrelated to the number of vanishing waveletmoments.The kth moment of a wavelet $ ( t ) is given by(8.147)Using the property (2.40) of the Fourier transform, the moments can also beexpressed as(8.148)Thus, if Q ( w ) has NQ zeros at W = 0, the wavelet has Nq, vanishing moments,that is00tk $(t)dt = 0 for k = 0, 1 , .
. . , Nq, - 1.(8.149)Clearly, the inner product of an analysis wavelet q ( t )having NG vanishingmoments with a signalcNq -1x(t)=aktkk=Ois zero, and, consequently, all wavelet coefficients are zero. Thus, polynomialsignals of order Nq - 1 are solely represented by the lowpass component, thatis, by the coefficients of the scaling function.244TransformChapter 8. WaveletThe number of vanishing moments is easily controlled when constructingwavelets from filter banks. In order to see this, let us recall equation (8.133):NG is given by the number of zeros of Hl(ej") at W = 0, or, equivalently,by the number of zeros of H l ( z ) at z = 1.
Note that according to (6.22),H1 ( z ) is a modulated version of the synthesis lowpass Go(z),so that we mayalternatively say that Nq is given by the number of zeros of Go(z) at z = -1.Similarly, the number of vanishing moments of the synthesis wavelet is equalto the number of zeros of the analysis lowpass at z = -1.The discrete-time filters also have vanishing moments and correspondingapproximation properties for discrete-time polynomial signals. For the kthderivative of(ej") =h1 ( n )e-jwn(8.151)Cnwe get(8.152)From this expression we see that if H l ( z ) has Nq zeros at z = 1, then h l ( n )has NG vanishing moments in the discrete-time sense:E n e hl(n) = o for IC = 0 , 1 , .
. . , N @- 1.(8.153)nThis means that sampled polynomial signals of order Nq - 1 are solelyrepresented by the lowpass component.8.6.6RegularityIn Figures 8.13 and 8.14 we saw that different filters may have completelydifferent convergence properties. Typically, one prefers smooth functions r$(t),which should possibly have several continuous derivatives.Daubechies deriveda test that can check the regularity and thus the convergence of the productin (8.125) [34]. Assuming that Go(.) has N zeros at z = -1, Go(.) can bewritten asN1 2-1GO(.) = fi (7S ( z ) . ) (8.154)+Note that N 2 1 because of (8.137). Further note that S(l) = 1 becauseof (8.135).
Pointwiseconvergence of the functions zi(t) defined in (8.128)2458.6. Wavelets from Filter Bankstowards a continuous function zoo@)= $(t) is guaranteed ifsup IS(ej')l<(8.155)05w527rClearly, if Go(z) hasno zero at z = -1, then (8.155) cannotbe satisfiedbecause S(l)= 1.If N is larger than theminimum number that is required to satisfy (8.155),then the function $(t)will also have continuous derivatives. Precisely, $(t)ism-times continuously differentiable if(8.156)Regularity isonly associated with the lowpass filters gO(n) and ho(n),respectively.
Given a continuous function $(t),the function I+!I(t)according to(8.129) will be continuous for any sequence g I ( n ) .HSlder Regularity. Rioulintroducedtheconceptof Holder regularity,which can be expressedas follows: if a scaling function is m-times continuouslydifferentiable anditsmth derivativeisHolder continuous of order a ,then its regularityis T = m a [125]. The Holder exponent a is the maximuma for whichI~'"'(t).)I171avt,.(8.157)++c8.6.7 Wavelets with Finite Supportand g 1 ( n ) are FIR filters, then the resulting scaling functions andwavelets have finite support [34].
The proofis straightforward. One merelyhas to consider the iteration (8.128) with the L coefficients go(O), . . . ,g o ( L - 1)while assuming that x i ( t ) is restricted to the interval [0, L - l]:If go(.)cL-l%+l( t )= Jzg o ( n ) Xi(2t - n).(8.158)n=OThen, all recursively constructed functions are restrictedto 0 5 2t -n 5 L - 1.Since the convergence is unique, xm(t) = $(t) is restricted to [0, L - l] forany arbitrary zo(t).The fact that the supportis known can be exploited to calculate the valuesof $(t)at the times t,, = n2rn. This, again,is based on the two-scale relation(8.119):$(t)= hg o ( l ) $(2t - .e).(8.159)ce246TransformChapter 8. WaveletLet us assume that the initial values $(n) are known.
By writing (8.159) as4($)=-JzC g o ( t ) $@- 4,$($l=-JzCgo(t)ee$(g -l),(8.160)we realize that we obtaintheintermediate values at each iterationstep.However, so far we only know the values $(O) = 0 and $ ( L ) = 0. The initialvalues required can be determinedby exploiting the fact that theinitial valuesremain unchanged during the iteration (8.160). Withm = [ $ P ) , .
. . , $ ( L - l)]'(8.161)m=M.m(8.162)we getaccording to (8.160), where the L-1X L-1 matrix M is given by[M]ij:= -Jz go(2i - j ) .(8.163)Recalling (8.140) it becomesobvious that we obtain the initial values bydetermining the right eigenvector m of M which belongs to the eigenvalue 1.Note. We concludefrom (8.135) and (8.138) that the sum of the evencoefficients equals the sum of the odd coefficients:(8.164)Since the columns of M containeither all even coefficients go(2n)or allodd coefficients go(2n l), the sum of all elements of the columns of Mis one. Thus, conditions (8.135) and (8.138) guarantee the existence of a lefteigenvector [l,1,.
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