Tài liệu Digital Signal Processing Handbook P35 ppt

Cormac Herley. “Wavelets and Filter Banks.” 2000 CRC Press LLC. <http://www.engnetbase.com>. WaveletsandFilterBanks CormacHerley HewlettPackardLaboratories 35.1FilterBanksandWavelets DerivingContinuous-TimeBasesFromDiscrete-TimeOnes • Two-ChannelFilterBanksandWavelets • StructureofTwo- ChannelFilterBanks • PuttingthePiecesTogether References 35.1 FilterBanksandWavelets Themethodsofdesigningbasesthatwewillemploydrawonideasfirstusedintheconstructionof multiratefilterbanks.Theideaofsuchsystemsistotakeaninputsystemandsplititintosubsequences usingbanksoffilters.Thissimplestcaseinvolvessplittingintojusttwopartsusingastructuresuch asthatshowninFig.35.1.Thistechniquehasalonghistoryofuseintheareaofsubbandcoding: firstofspeech[1,2]andmorerecentlyofimages[3,4].Infact,themostsuccessfulimagecoding schemesarebasedonfilterbankexpansions[5,6,7].Recenttextsonthesubjectare[8,9,10].We willconsideronlythetwo-channelcaseinthissection.If ˆ X(z)=X(z),thenthefilterbankhasthe perfectreconstructionproperty. FIGURE35.1:Maximallydecimatedtwo-channelmultiratefilterbank. Itiseasilyshownthattheoutput ˆ X(z)oftheoverallanalysis/synthesissystemisgivenby: ˆ X(z)= 1 2 [G 0 (z)G 1 (z)]  H 0 (z)ψ H 0 (−z) H 1 (z)ψ H 1 (−z)  X(z) X(−z)  (35.1) = 1 2 [H 0 (z)G 0 (z)+H 1 (z)G 1 (z)]·X(z) c  1999byCRCPressLLC + 1 2 [H 0 (−z)G 0 (z) + H 1 (−z)G 1 (z)]·X(−z). Call the above 2 × 2 matrix H m (z). This gives that the unique choice for the synthesis filters is  G 0 (z) G 1 (z)  =  H 0 (z) H 0 (−z) H 1 (z) H 1 (−z)  −1 ·  2 0  = 2  m (z)  H 1 (−z) −H 0 (−z)  , (35.2) where  m (z) = detH m (z). Ifweobservethat m (z) =− m (−z) anddefineP(z) = 2·H 0 (z)H 1 (−z)/ m (z) = H 0 (z)G 0 (z), it follows from (35.2) that G 1 (z)H 1 (z) = 2 ·H 1 (z)H 0 (−z)/ m (−z) = P(−z). We can then write that the necessary and sufficient condition for perfect reconstruction ( 35.1) is: P(z)+ P(−z) = 2. (35.3) Sincethisconditionplaysan important roleinwhat follows, wewillrefertoanyfunction having this propertyas valid. The implicationof thispropert y isthat allbut one ofthe even-indexedcoefficients of P(z)are zero. That is P(z)+ P(−z) =  n (p(n)z −n + p(n)(−z) −n ) =  n 2 ·p(2n)z −(2n+1) . For this to satisfy (35.3) requires p(2n) = δ n ; thus, one of the polyphase components of P(z)must be the unit sample. By polyphasecomponents we mean the set of even-indexed samples,and the set of the odd-indexed samples. Such a function is illustrated in Fig. 35.2(a). FIGURE 35.2: Zeros of the correlation functions. (a) Autocorrelation H 0 (−z)H 0 (z −1 ). (b) Cross- correlation H 0 (−z)H 1 (z −1 ). Constructingsuchafunctionisnotdifficult. Ingeneral,however,wewillwishtoimposeadditional constraints on the filter banks. So, P(z)w ill have to satisfy other constraints in addition to (35.3). Observe that as a consequence of (35.2) G 0 (z)H 1 (z), i.e., the cross-correlation of g 1 (n) and the time-reversed filter h 0 (−n), and G 1 (z)H 0 (z), the cross-correlation of g 1 (n) and h 0 (−n), have only odd-indexed coefficients, just as for the function in Fig. 35.2(b), that is: <g 0 (n), h 1 (2k −n) > = 0, (35.4) c  1999 by CRC Press LLC <g 1 (n), h 0 (2k −n) > = 0, (35.5) (note the time reversal in the inner product). Define now the matrix H 0 as H 0 =       . . . . . . . . . . . . . . . . . . . . . . . . h 0 (L − 1)h 0 (L − 2) ··· ··· h 0 (0) 00 00h 0 (L − 1) ··· h 0 (2)h 0 (1)h 0 (0) . . . . . . . . . . . . . . . . . . . . . . . .       . (35.6) whichhasasitskthrowtheelementsof the sequenceh 0 (2k −n). Pre-multiplyingbyH 0 corresponds to filtering by H 0 (z) followed by subsampling by a factor of 2. Also define G T 0 =       . . . . . . . . . . . . . . . . . . . . . . . . g 0 (0)g 0 (1) ··· ··· g 0 (L − 1) 00 00g 0 (0) ··· g 0 (L − 3)g 0 (L − 2)g 0 (L − 1) . . . . . . . . . . . . . . . . . . . . . . . .       , (35.7) so G 0 has as its kth column the elements of the sequence g 0 (n − 2k). Define H 1 by replacing the coefficients ofh 0 (n) with those of h 1 (n) in (35.6) and G 1 by replacing the coefficients of g 0 (n) with those of g 1 (n) in (35.7). Wefindthat (35.4)givesthatallrowsofH 1 areorthogonaltoallcolumnsofG 0 . Similarlywefind, from (35.5), that all of the columns of G 1 are orthogonal to the rows of H 0 . So, in matrix notation: H 0 G 1 = 0 = H 1 G 0 . (35.8) Now P(z) = G 0 (z)H 0 (z) = z −1 H 0 (z)H 1 (−z) and P(−z) = G 1 (z)H 1 (z) are b oth valid and have theform given inFig. 35.2 (a). Hence, theimpulse responses ofg i (n) andh i (n) are orthogonal with respect to even shifts <g i (n), h i (2l −n) > = δ l . (35.9) In operator notation: H 0 G 0 = I = H 1 G 1 . (35.10) Sincewehaveaperfectreconstructionsystemweget: G 0 H 0 + G 1 H 1 = I. (35.11) Of course (35.11) indicates that no nonzero vector can lie in the column nullspaces of both G 0 and G 1 . Notethat (35.10) implies thatG 0 H 0 and G 1 H 1 areeachprojections(sinceG i H i G i H i = G i H i ). They project onto subspaces which are not, in general, orthogonal (since the operators are not self- adjoint). Because of (35.4), (35.5),and (35.9)the analysis/synthesissystem is termed biorthogonal. If we interleave the rows of H 0 and H 1 , much as was done in the orthogonal case, and form again a block Toeplitz matrix A =           . . . . . . . . . . . . . . . . . . . . . . . . h 0 (L − 1)h 0 (L − 2) ··· ··· h 0 (0) 00 h 1 (L − 1)h 1 (L − 2) ··· ··· h 1 (0) 00 00h 0 (L − 1) ··· h 0 (2)h 0 (1)h 0 (0) 00h 1 (L − 1) ··· h 1 (2)h 1 (1)h 1 (0) . . . . . . . . . . . . . . . . . . . . . . . .           , (35.12) c  1999 by CRC Press LLC wefind thatthe rowsof A form a basisfor l 2 (Z).IfweformBby interleaving the columnsof G 0 and G 1 ,wefind B · A = I. In the special case where we have a unitary solution, one finds: G 0 = H T 0 and G 1 = H T 1 , and (35.8) gives that we have projections onto subspaces which are mutually orthogonal. The system then simplifies to the orthogonal case, where B = A −1 = A T . A point that we wish to emphasize is that in theconditions for perfect reconstruction, (35.2) and (35.3), the filters H 0 (z) and G 0 (z) are related via their product P(z). It is the choice of the function P(z)and the factorization taken that determines the properties of the filter bank. We conclude the introduction with a proposition that sums up the foregoing. PROPOSITION35.1 Todesignatwo-channelperfectreconstruction filter bank, itisnecessaryand sufficient to finda P(z)satisfying (35.3),factorit P(z) = G 0 (z)H 0 (z) andassign thefilters as given in (35.2). 35.1.1 Deriving Continuous-Time Bases From Discrete-Time Ones We have seen that the construction of bases from discrete-time signals can be accomplished easily by using a perfect reconstruction filter bank as the basic building block. This gives us bases that have a certain structure, and for which the analysis and synthesis can be efficiently performed. The design of bases for continuous-time signals appears more difficult. However, it works out that we can mimic many of the ideas used in the discrete-time case, when we go about the construction of continuous-time bases. In fact, there is a very close correspondence between the discrete-time bases generated by two- channel filter banks, and dyadic wavelet bases. These are continuous-time bases formed by the stretches and translates of a single function, where the stretches are integer powers of two: {ψ jk (x) = 2 −j/2 ψ(2 −j x −k), j, k, ∈ Z} (35.13) This relation has been thoroughly explored in [11, 12]. To be precise, a basis of the form in (35.13) necessarily implies the existence of an underly ing two-channel filter bank. Conversely, a two-channel filter bank can be used to generate a basis as in (35.13) provided that the lowpass filter H 0 (z) is regular. It is not our intention to go into the details of this connection, but the generation of wavelets from filter banks goes briefly as follows: Considering the logarithmic tree of discrete-time filters in Fig. 35.3, one notices that the lower branch is a cascade of filters H 0 (z) followed by subsampling by 2. It is easily shown [12], that the cascade of i blocks of filtering operations, followed by subsampling by 2, is equivalent to a filter H (i) 0 (z) with z-transform: H (i) 0 (z) = i−1  l=0 H 0 (z 2 l ), i = 1, 2 ···, (35.14) followed by subsampling by 2 i . We define H (0) 0 (z) = 1 to initialize the recursion. Now, in addition to the discrete-time filter, consider the function f (i) (x) which is piecewise constant on intervals of length 1/2 i , and equal to: f (i) (x) = 2 i/2 · h (i) 0 (n), n/2 i ≤ x<(n+ 1)/2 i . (35.15) Notethat the normalizationby2 i/2 ensuresthatif  (h (i) 0 (n)) 2 = 1 then  (f (i) (x)) 2 dx = 1 as well. Also, it can be checked that h (i) 0  2 = 1 when h (i−1) 0  2 = 1. The relation between the sequence c  1999 by CRC Press LLC H (i) 0 (z) and the function f (i) (x) is clarified in Fig. 35.3, where the first three iterations of each is shown for the simple case of a filter of length 4. FIGURE35.3: Iterationsofthediscrete-timefilter(35.14)andthecontinuous-timefunction(35.15) for the case of a length-4 filterH 0 (z). The length ofthe filterH (i) 0 (z) increases withoutbound, while the function f (i) (x) actually has bounded support. We are going touse thesequenceof functionsf (i) (x) toconvergetothe scalingfunction φ(x)of a waveletbasis. Hence,a fundamentalquestionis tofind out whetherand towhat the functionf (i) (x) converge s as i →∞. First assume that the filter H 0 (z) has a zero at the half sampling frequency, or H 0 (e jπ ) = 0. This together with the fact that the filter impulse response is orthogonal to its even translates is equivalent to  h 0 (n) = H 0 (1) = √ 2. Define M 0 (z) = 1/ √ 2 · H 0 (z), that is M 0 (1) = 1.NowfactorM 0 (z) intoitsrootsatπ (thereisatleastonebyassumption)andaremainder polynomial K(z), in the following way: M 0 (z) =[(1 + z −1 )/2] N K(z). Note that K(1) = 1 from the definitions. Now call B the supremum of |K(z)| on the unit circle: B = sup ω∈[0,2π ] |K(e jω )|. Then the following result from [11] holds: c  1999 by CRC Press LLC PROPOSITION35.2 [Daubechies 1988] If B<2 N−1 , and ∞  n=−∞ |k(n)| 2 |n|  < ∞, for some >0, (35.16) thenthe piecewise constantfunction f (i) (x) definedin (35.15) convergespointwise toa continuous function f (∞) (x). This is a sufficient condition to ensure pointw ise convergence to a continuous function, and can beusedasa simple test. We shallrefertoanyfilterforwhichtheinfiniteproductconvergesasregular. If we indeed have convergence, then we define f (∞) (x) = φ(x) as the analysis scaling function, and ψ(x) = 2 −1/2  h 1 (n)φ(2x − n), (35.17) as the analysis wavelet. It can be shown that if the filters h 0 (n) and h 1 (n) arefromaperfectrecon- struction filter bank, then (35.13) indeed forms a continuous-time basis. In a similar way we examine the cascade of i blocks of the synthesis filter g 0 (n) G (i) 0 (z) = i−1  l=0 G 0 (z 2 l ), i = 1, 2 ···. (35.18) Again, define G (0) 0 (z) = 1 to initialize the recursion, and normalize G 0 (1) = 1. From this define a function which is piecewise constant on intervals of length 1/2 i : ˇ f (i) (x) = 2 i/2 · g (i) 0 (−n), n/2 i ≤ x<(n+ 1)/2 i . (35.19) We call the limit ˇ f (∞) (x), if it exists, ˇ φ(x) the synthesis scaling function, and we find ˇ φ(x) = 2 1/2 · L−1  n=0 g 0 (−n) · ˇ φ(2x −n) (35.20) ˇ ψ(x) = 2 1/2 · L−1  n=0 g 1 (−n) · ˇ φ(2x −n). (35.21) The biorthogonality properties of the analysis and synthesis continuous-time functions follow from the corresponding properties of the discrete-time ones. That is, (35.9) leads to < ˇ φ(x),φ(x − k) > = δ k . (35.22) and < ˇ ψ(x), ψ(x −k) > = δ k . (35.23) Similarly < ˇ φ(x),ψ(x − k) > = 0 (35.24) < ˇ ψ(x), φ(x −k) > = 0, (35.25) c  1999 by CRC Press LLC come from (35.4) and (35.5), respectively. We have shown that the conditions for perfect reconstruction on the filter coefficients lead to functions that have the biorthogonality properties as shown above. Orthogonality across scales is also easily verified: < ˇ ψ(2 j x),ψ(2 i x −k) > = δ i−j δ k . Thus, theset {ψ(2 j x), ˇ ψ(2 i x −k),i,j,k ∈ Z}is biorthogonal. Thatit is completecan be verified as in the orthogonal case [13]. Hence, any function from L 2 (R) can be written: f(x) =  j  l < f (x), 2 −j/2 ψ(2 j x −l) > 2 −j/2 ˇ ψ(2 j x −l). Note that ψ(x) and ˇ ψ(x) play interchangeable roles. 35.1.2 Two-Channel Filter Banks and Wavelets We have seen that the design of discrete-time bases is not difficult: using two-channel filter banks as thebasicbuilding block theycan be easilyderived. We alsoknowthat,using (35.15)and (35.19),we cangenerate continuous-timebases quite easilyas well. Ifwe were just interestedin the construction of bases, with no further requirements, we could stop here. However, for applications such as com- pression, we will often be interested in other propertiesof the basis functions, for example, whether or not they have any symmetry or finite support, and whether or not the basis is an orthonormal one. We examinethesethreestructuralproperties fortheremainderofthissection. Chapter36deals with the design of the filters. Chapter 37 deals with time-varying filter banks, where the filters used, or the tree structure employing them, varies over time. Chapter 38 deals with the case of Lapped Transforms, a veryimportantclass of multirate filter banks that have achieved considerable success. From the filter bank point of view, the properties we are most interested in are the following: • Orthogonality: <h 0 (n), h 0 (n + 2k) > = δ k = <h 1 (n), h 1 (n + 2k) >, (35.26) <h 0 (n), h 1 (n + 2k) > = 0. (35.27) • Linearphase: H 0 (z), H 1 (z), G 0 (z), and G 1 (z) are all linear phase filters. • Finite support: H 0 (z), H 1 (z), G 0 (z), and G 1 (z) are all FIR filters. The reason for our interest is twofold. First, these properties are possibly of value in perfect reconstruction filter banks used in subband coding schemes. For example, orthogonality implies that the quantization noise in the two channels will be independent; linear phase is possibly of interestinverylowbit-ratecodingof images, andFIRfiltershavetheadvantageofhavingvery simple low-complexity implementations. Second, these properties are carried over to the wavelets that are generated. So, if we design a filter bank with a certain set of properties, then the continuous-time basis that it generates will also have these properties. PROPOSITION35.3 If the filters belong to an orthogonal filter bank, we shall have < φ(x), φ(x +k) > = δ k =< ψ(x), ψ(x +k) >, < φ(x), ψ(x +k) > = 0. c  1999 by CRC Press LLC PROOF35.1 Fromthedefinition (35.15)f (0) (x) isjusttheindicatorfunctionontheinterval[0, 1); so we immediately get orthogonality at the 0th level, that is: <f (0) (x − l), f (0) (x − k) > = δ kl . Now we assume orthogonality at the ith level: <f (i) (x −l), f (i) (x −k) > = δ kl , (35.28) and prove that this implies orthogonality at the (i +1)st level: <f (i+1) (x −l), f (i+1) (x −k) > = 2  n  m h 0 (n)h 0 (m) <f (i) (2x −2l − n), f (i) (2x −2k − m) > δ n+2l−2k−m 2 =  n h 0 (n)h 0 (n + 2l −2k) = δ kl . Hence, by induction (35.28) holds for all i. So in the limit i →∞: <φ(x−l),φ(x − k) > = δ kl . (35.29) The orthogonal casegives considerable simplification, both in thediscrete-time and continuous- time cases. PROPOSITION 35.4 If the filters belong to an FIR filter bank, then φ(x), ψ(x), ˇ φ(x), and ˇ ψ(x) will have support on some finite interval. PROOF 35.2 ThefiltersH (i) 0 (z) andG (i) 0 (z) definedin (35.14)haverespectivelengths(2 i −1)(L a − 1) +1 and (2 i −1)(L s −1) +1 whereL a andL s arethelengths of H 0 (z) and G 0 (z).Hence,f (i) (x) in (35.15) is supported on the interval [0,L a − 1) and ˇ f (i) (x) on the interval [0,L s − 1). This holds ∀ i; hence, in the limit i →∞this gives the support of the scaling functions φ(x) and ˇ φ(x). That ψ(x) and ˇ ψ(x) have bounded support follow from (35.20) and (35.21). PROPOSITION35.5 If the filters belong to a linear phase filter bank, then φ(x), ψ(x), ˇ φ(x), and ˇ ψ(x) will be symmetric or antisymmetric. PROOF 35.3 The filter H (i) 0 (z) will have linear phase if H 0 (z) does. If H (i) 0 (z) has length (2 i − 1)(L a −1)+1,thepointof symmetry is(2 i −1)(L a −1)/2 whichneed not beaninteger. Thepoint of symmetry for f (i) (x) will then be [(2 i − 1)(L a − 1) + 1]/2 i+1 or [(2 i − 1)(L a − 1) + 2]/2 i+1 . Ineither case,by takingthe limiti →∞we findthat φ(x)is symmetric aboutthe point(L a −1)/2 and similarly for the other cases. Thushavingestablishedtherelationbetweenwaveletsandfilterbankswecanexaminethestructure of filter banks in detail, and afterward usethem to generate waveletsas described above. It shouldbe emphasized thatwe are speaking of the two-channel, one-dimensional case. Multidimensional filter banks are a large subject in their own right [8, 10]. c  1999 by CRC Press LLC 35.1.3 Structure of Two-Channel Filter Banks We saw already that it is the choice of the function P(z)and the factorization taken that determines the propertiesof thefilterbank. Interms ofP(z),we givenecessaryand sufficientconditionsfor the three properties mentioned above: • Orthogonality: P(z)isan autocorrelation, and H 0 (z) and G 0 (z) areits spectral factors. • Linearphase: P(z)is linear phase, and H 0 (z) and G 0 (z) are its linear phase factors. • Finite support: P(z)is FIR, and H 0 (z) and G 0 (z) are its FIR factors. Obviouslythefactorizationisnotuniqueinanyofthecasesabove. TheFIRcasehasbeenexamined in detail in [11, 12, 14, 15, 16] and the linear phase case in [12, 15, 17]. In the rest of this paper we will present new results on the orthogonal case, but we shall also review the solutionsthat explicitly satisfy simultaneous constraints. PROPOSITION 35.6 To have an orthogonal filter bank it is necessary and sufficient that P(z)be an autocorrelation, and that H 0 (z) and G 0 (z) be its spectral factors. PROPOSITION35.7 To have a linear phase filter bank it is necessary and sufficient that P(z)be a linear phase, and that H 0 (z) and G 0 (z) be its linear phase factors. PROPOSITION35.8 To have an FIR filter bank itis necessary and sufficient thatP(z)be FIR, and that H 0 (z) and G 0 (z) be its FIR factors. Proofs can be found in [18]. Having seen that the design problem can be considered in terms of P(z)and its factorizations, we consider the three conditions of interest from this point of view. Orthogonality In the case where the filter bank is to be orthogonal, we can obtain a complete constructive characterization of the solutions, as given by the following theorem, taken from [18]. THEOREM 35.1 All orthogonal rational two channel filter banks can be for med as follows: 1. Choosing an arbitrary polynomial R(z), form: P(z) = 2 ·R(z)R(z −1 ) R(z)R(z −1 ) + R(−z)R(−z −1 ) , 2. factor as P(z) = H(z)H(z −1 ), 3. form the filter H 0 (z) = A 0 (z)H (z),whereA 0 (z) is an arbitrary allpass, 4. choose H 1 (z) = z 2k−1 H 0 (−z −1 )A 1 (z 2 ),whereA 1 (z) is again an arbitrary allpass, 5. choose G 0 (z) = H 0 (z −1 ), and G 1 (z) =−H 1 (z −1 ). Foraproof,see[18, 19]. c  1999 by CRC Press LLC [...]... anata, Z., The role of lossless systems in modern digital signal g processing, IEEE Trans Education, 32, 181–197, Aug 1989 Special issue on Circuits and Systems e [27] Meyer, Y., Ondelettes, vol 1 of Ondelettes et Op´rateurs, Hermann, Paris, 1990 [28] Lawton, W., Application of complex-valued wavelet transforms to subband decomposition, IEEE Trans on Signal Proc., submitted, 1992 c 1999 by CRC Press... 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