Tài liệu 35 Wavelets and Filter Banks docx

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)ψH 0 (−z) H 1 (z)ψH 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) = det H m (z). Ifweobservethat  m (z) =− m (−z)anddefine P(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) Since this condition plays an important role in what follows, we will refer to any function having this property as valid. The implication of this property is that all but one of the even-indexed coefficients 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 polyphase components 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 ). Constructing such a function is not difficult. In general, however, we will wish to impose additional constraints on the filter banks. So, P(z)will 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) which has as its kth row the elements of the sequence h 0 (2k− n). Pre-multiplying by H 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 of h 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). We find that (35.4) gives that all rows of H 1 are orthogonal to all columns of G 0 . Similarly we find, 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 both valid and have the form given in Fig. 35.2 (a). Hence, the impulse responses of g i (n) and h 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 . Note that (35.10) implies that G 0 H 0 and G 1 H 1 are each projections (since G 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/synthesis system 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 we find that the rows of A form a basis for l 2 (Z).IfweformB by interleaving the columns of 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 the conditions 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 To design a two-channel perfect reconstruction filter bank, it is necessary and sufficient to find a P(z)satisfying (35.3), factor it P(z)= G 0 (z)H 0 (z) and assign the filters 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 underlying 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) Note that the normalization by 2 i/2 ensures that if  (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. FIGURE 35.3: Iterations of the discrete-timefilter ( 35.14) and the continuous-time function ( 35.15) for the case of a length-4 filter H 0 (z). The length of the filter H (i) 0 (z) increases without bound, while the function f (i) (x) actually has bounded support. We are going to use the sequence of functions f (i) (x) to converge to the scaling function φ(x)of a wavelet basis. Hence, a fundamental question is to find out whether and to what the function f (i) (x) converges 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) into its roots at π (there is at least one by assumption) and a remainder 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 PROPOSITION 35.2 [Daubechies 1988] If B<2 N−1 , and ∞  n=−∞ |k(n)| 2 |n|  < ∞, for some >0, (35.16) then the piecewise constant function f (i) (x) defined in (35.15) converges pointwise to a continuous function f (∞) (x). This is a sufficient condition to ensure pointwise convergence to a continuous function, and can be used as a simple test. We shall refer to any filter for which the infinite product converges as regular. 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, the set {ψ(2 j x), ˇ ψ(2 i x − k),i,j,k ∈ Z} is biorthogonal. That it is complete can 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 the basic building block they can be easily derived. We also know that, using (35.15) and (35.19), we can generate continuous-time bases quite easily as well. If we were just interested in 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 properties of 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 examine these three structural properties for the remainder of this section. Chapter 36 deals 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 very important class 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) • Linear phase: 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 interest in very low bit-rate coding of images, and FIR filters have the advantage of having very 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. PROPOSITION 35.3 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 Fromthe definition (35.15) f (0) (x)is justthe indicatorfunction on the interval[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 case gives considerable simplification, both in the discrete-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. PROOF35.2 The filters H (i) 0 (z) and G (i) 0 (z) defined in (35.14) have respectivelengths (2 i −1)(L a − 1)+1 and (2 i −1)(L s −1)+ 1 where L a and L s are the lengths 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). PROPOSITION 35.5 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, the point of symmetry is (2 i −1)(L a −1)/2 which need not be an integer. The point 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 . In either case, by taking the limit i →∞we find that φ(x)is symmetric about the point (L a − 1)/2 and similarly for the other cases. Thushavingestablishedtherelationbetween wavelets and filter banks we can examine the structure of filter banks in detail, and afterward use them to generate wavelets as described above. It should be emphasized that we 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 properties of the filter bank. In terms of P(z), we give necessary and sufficient conditions for the three properties mentioned above: • Orthogonality: P(z)is an autocorrelation, and H 0 (z) and G 0 (z) are its spectral factors. • Linear phase: 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. Obviously thefactorization is not unique in any of the cases above. The FIR case has been examined 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 solutions that 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. PROPOSITION 35.7 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. PROPOSITION 35.8 To have an FIR filter bank it is necessary and sufficient that P(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 formed 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 [...]... supported wavelets, Communications on Pure and Applied Mathematics, XLI, 909–996, 1988 [12] Vetterli, M and Herley, C., Wavelets and filter banks: theory and design, IEEE Trans on Signal Proc., 40, 2207–2232, Sept 1992 [13] Cohen, A., Daubechies, I and Feauveau, J.-C., Biorthogonal bases of compactly supported wavelets, Commun on Pure and Applied Mathematics, 45, 485–560, 1992 [14] Smith, M.J.T and Barnwell... Two-channel perfect-reconstruction FIR QMF structures which yield linear-phase analysis and synthesis filters, IEEE Trans Acoust., Speech, Signal Proc., 37, 676–690, May 1989 [18] Herley, C and Vetterli, M., Wavelets and recursive filter banks, IEEE Trans on Signal Proc., 41, 2536–2556, Aug 1993 [19] Herley, C., Wavelets and Filter Banks, Ph.D thesis, Columbia University, New York, April 1993 Available by anonymous... the following shorthand notation to list the coefficients of a causal FIR sequence: N−1 an z−n = (a0 , a1 , a2 , · · · aN −1 ) n=0 So, using the description of the filters in Theorem 35. 1, with the simplest case A0 (z) = A1 (z) = 1 and k = 0 we find: H0 (z) = H1 (z) = G0 (z) = (1 + 7z−1 + 21z−2 + 35z−3 + 35z−4 + 21z−5 + 7z−6 + z−7 ) √ 2 · (1 + 21z−2 + 35z−4 + 7z−6 ) 1 + 21z2 − 35z3 + 35z4 − 21z5 + 7z6 −... Xiong, Z., Ramchandran, K and Orchard, M.T., Wavelet packet image coding using spacefrequency quantization, IEEE Trans on Image Proc., submitted, 1996 [8] Vaidyanathan, P.P., Multirate Systems and Filter Banks, Prentice-Hall, Englewood Cliffs, NJ, 1992 [9] Malvar, H.S., Signal Processing with Lapped Transforms, Artech House, 1992 [10] Vetterli, M and Kovacevic, J., Wavelet and Subband Coding, Prentice-Hall,... reconstruction filter banks for tree structured subband coders, in Proc IEEE Intl Conf ASSP, San Diego, CA, pp 27.1.1– 27.1.4, March 1984 [23] Mintzer, F., Filters for distortion-free two-band multirate filter banks, IEEE Trans Acoust., Speech, Signal Proc., 33, 626–630, June 1985 [24] Vaidyanathan, P.P and Hoang, P.-Q., Lattice structures for optimal design and robust implementation of two-band perfect reconstruction... 35z3 + 35z4 − 21z5 + 7z6 − z7 ) (1 − 7z z−1 √ 2 · (1 + 21z2 + 35z4 + 7z6 ) H0 (z−1 ) G1 (z) = H1 (z−1 ) In the notation of Proposition 35. 2, B = 8 < 26 so that for this choice of H0 (z) the left-hand side of (35. 15) converges to a continuous function The wavelet, scaling function, and their spectra are shown in Fig 35. 4 Finite Impulse Response and Symmetric Solutions In the case where the filters are to... and FIR; i.e., they were in A ∩ B A constructive parametrization of A ∩ B was given in [24] The construction c 1999 by CRC Press LLC and characterization of examples which converge to wavelets was first done in [11] Filter banks with FIR linear phase filters (i.e., A ∩ C) were first given in [15], and also studied in terms of lattices in [17, 25] The construction of wavelet examples is given in [13] and. .. reconstruction for tree-structured subband coders, IEEE Trans Acoust., Speech, Signal Proc., 34, 434–441, June 1986 [15] Vetterli, M., Filter banks allowing perfect reconstruction, Signal Proc., 10(3), 219–244, 1986 [16] Vaidyanathan, P.P., Multirate digital filters, filter banks, polyphase networks, and applications: a tutorial, Proc IEEE, 78, 56–93, Jan 1990 [17] Nguyen, T.Q and Vaidyanathan, P.P., Two-channel... N matrix, k2N = (k0 , · · · , k(k−1) ), and e2N is the length k vector (0, 0, · · · , 1) Having found the coefficients of K2N (z), we factor it into linear phase components and then regroup these factors of K2N (z) and the 2N zeros at z = −1 to form two filters: H0 (z) and H1 (−z), both of which are to be regular c 1999 by CRC Press LLC FIGURE 35. 5: Biorthogonal wavelets generated by filters of length... wavelet examples is given in [13] and [12] Filter banks, which are linear phase and orthogonal, were constructed in Chapter 36 and were presented in [18] That there exist only trivial solutions which are linear phase, orthogonal and FIR is indicated by the intersection A ∩ B ∩ C; the only solutions are two tap filters [11, 12, 26] It warrants emphasis that Fig 35. 6 illustrates the filter bank solutions; . HewlettPackardLaboratories 35. 1FilterBanksandWavelets DerivingContinuous-TimeBasesFromDiscrete-TimeOnes • Two-ChannelFilterBanksandWavelets • StructureofTwo- ChannelFilterBanks. Herley. Wavelets and Filter Banks. ” 2000 CRC Press LLC. <http://www.engnetbase.com>. WaveletsandFilterBanks CormacHerley HewlettPackardLaboratories 35. 1FilterBanksandWavelets