Parametrized Biorthogonal Wavelets and FIR Filter

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Parametrized Biorthogonal Wavelets and FIR Filter Bank Design with Gr¨obner Bases MinHsuan Peng and N. K. Bose

Abstract This paper builds upon the recent results of Regensburger and Scherzer on parametrization of orthonormal wavelets by discrete moments (and, therefore, also continuous moments) of scaling function followed by the solution of a parametrized set of polynomial equations in the FIR filter coefficients using Gr¨obner bases. First, parametrization of orthonormal filters with two discrete moments as parameters, instead of just one is considered followed by the generalization of the results to the case of biorthogonal wavelets. A characterization theorem for the continuous moments of the scaling function and its dual is presented and proved for biorthogonal wavelets and the result of Regensburger and Scherzer in the orthogonal case is shown to emerge as a special case of this. Like in Regensburger’s subsequent work, complete solution to parametrized filters for both the two-parameter orthogonal case and the biorthogonal case (especially for the practically important linear phase biorthogonal class) for filters of varying lengths are presented. It is shown that for the interesting example in image compression considered by Regensburger, better performance results are obtained with the two-parameter parametrization considered in this paper.

Index Terms – Parametrization of Orthogonal and biorthogonal wavelets, Gr¨obner bases, vanishing moments of wavelets, image compression. I. I NTRODUCTION In a recent paper, Regensburger and Scherzer [1] established explicit and not recursive, as done earlier (see for example [2]) bijective relations between continuous and discrete moments of scaling functions associated with orthogonal wavelets by expressing the nth continuous moment as a polynomial (related to a class of distinguished polynomials, called Bell polynomials) of the first n discrete moments and vice versa. Subsequently, they parametrized the filter coefficients in the dilation equations generating the scaling equation, and compactly supported orthonormal wavelets with several vanishing moments, in terms of the discrete (and, consequently, the continuous) moments. By giving up certain vanishing moments to obtain additional degrees of freedom, they were able to generate a parametrized solution set in contrast to previous methods on symbolic computation of wavelets coefficients which calculated a finite number of solutions that also used algorithmic algebraic tools like Gr¨obner bases [3] [4, Chapter 4]. Subsequently, Regensburger [5] presented new simplified parametrizations and discussed computational aspects for up to ten filter coefficients and at least four vanishing moments with a single discrete moment as parameter. These results in [1] and [5] are significant because previous approaches to parametrization expressed the filter coefficients in terms of trigonometric functions (and not polynomials) which required the solving of transcendental constraints for the parameters to find wavelets with more than one vanishing moment. The algebraic and geometric structure of the space of compactly supported biorthogonal wavelets by factorization has been presented recently [6]. Construction of symmetric biorthogonal wavelets and smoothness estimates of such wavelets still remain largely unexplored. Such problems are addressed here through parametrization by discrete moments. It is known that vanishing moments of scaling and wavelet functions have different roles. Vanishing moments of the wavelet function is related to the wavelet orthogonal projection, while vanishing moments of the scaling function is connected to the wavelet sampling approximation inside the projection subspace [7]. Here, attention is given to both and solutions for the parametrized FIR (finite impulse response) polynomial filter are obtained by the symbolic computer algebra software SINGULAR used to construct Gr¨obner bases. In Section 2, the results of length ten filter in [1] and [5] for the single parameter case is very briefly reviewed as background. Then the two-parameter case for parametrization of filter coefficients for scaling function and orthogonal wavelets is considered. The counterpart of the results in [1] and [5] for the biorthogonal case developed in Section 3. Applications are considered in Section 4 and concluding remarks are summarized in Section 5. II. T WO PARAMETERS PARAMETRIZATION OF C OMPACTLY S UPPORTED O RTHOGONAL WAVELETS The maximum number p of vanishing moments Vl , l = 0, 1, ..., p − 1 of the wavelet ψ(x), Z Vl = xl ψ(x) = 0 Department of Electrical Engineering, The Pennsylvania State University, University Park, PA 16802, USA E-Mail: [email protected] This research was stimulated and initiated by the Special Semester on Grobner Bases supported by RICAM (the Radon Institute for Computational and Applied Mathematics, Austrian Academy of Science, Linz) and organized by RICAM and RISC (Research Institute for Symbolic Computation, Johannes Kepler University, Linz, Austria) under the scientific direction of Professor Bruno Buchberger.

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leads to a maximally flat FIR filter of length 2p that can be used to generate the Daubechies wavelet dbp. Therefore, a length ten filter generates the compactly supported orthonormal Daubechies wavelet db5, i. e. p = 5. By giving up the highest vanishing moment Vp−1 , one obtains one degree of freedom to parameterize the filter coefficients. Therefore, the remaining number of vanishing moments is four. Applying [1, Theorem 3.4], the even-indexed discrete moments mk , k = 0, 2, 4, ... of the associated scaling function can be determined by the odd-indexed discrete moments mk , k = 1, 3, 5, ... up to the number of vanishing moments of the wavelet. By this result, the discrete moments m1 , m2 , m3 , m4 for a length 10 filter, after giving up 1 vanishing moment are related as follows [1]: 1 2 m 2 1 3 m4 = − m41 + 2m1 m3 , 8 Using discrete moments m1 and m3 as parameters to parameterize a length ten filter with Gr¨obner bases, the first polynomial obtained is only in terms of m1 and m3 . If only one degree of freedom is chosen for parametrization as done in [5], then m1 and m3 are dependent. Here two degrees of freedom with two independent parameters are chosen for parametrization. In order to obtain two degrees of freedom, one needs to give up the two highest vanishing moments. However, the first vanishing moment V0 is a requirement for the resolution of identity in the theory of wavelets and cannot be eliminated. Therefore, for parametrization with two parameters (obtained by giving up the highest two vanishing moments) the minimum filter length is six. Therefore, two degrees of freedom for parametrization will be considered by giving up the highest two vanishing moments starting from FIR filters of length six up to length twelve in increments of two. m2 =

A. Setup: The linear system used P to parameterize the filter coefficients contains the equation of normalization (obtained from the scaling function φ(x) = hk φ(x − k)), N X

hk = 2,

(1)

k=0

the equation from discrete moments of scaling function, mn =

N X

hk k n ,

k=0

and the sum rule equation, obtained from vanishing moments of wavelet function ψ(x) = 2n−1 X

(−1)n−k (n − k)l hk = 0.

(2) P (−1)k hN −k ψ(x − k), (3)

k=0

By orthonormality property of scaling function,

Z φ(x)φ(x − l) = δ0,l ,

(4)

one can obtain the quadratic equations: N X

hk hk−2l = 2δ0,l , l = 0, ..., N − 1.

(5)

k=1−N

B. Length Six Case For parametrization with two parameters (for brevity, two-parametrization), give up one more vanishing moment in comparison to the one-parametrization case, which means that the only vanishing moment is the zeroth order moment Z ψ(x) = 0, and it is equivalent to (from sum rule defined in Eq.(3)) N X

(−1)k hk = 0.

k=0

(6)

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The following linear equations result from the normalization equation (1), the first sum rule (6) and two discrete moments m1 and m2 . Note that since there is only one vanishing moment, therefore, m2 is not determined by m1 . h0 + h1 + h2 + h3 + h4 + h5 h0 − h1 + h2 − h3 + h4 − h5 h1 + 2h2 + 3h3 + 4h4 + 5h5 h1 + 4h2 + 9h3 + 16h4 + 25h5

=2 =0 =m1 =m2

(7)

The quadratic equations are: h0 h2 + h1 h3 + h2 h4 + h3 h5 =0 h0 h4 + h1 h5 =0.

(8)

To solve for the six filter coefficients, parametrized by m1 and m2 , use SINGULAR to obtain the reduced Gr¨obner basis based on lexicographical order and ranking h5 > ... > h1 > h0 in the form of a triangular system of polynomial equations: 16384h40 + (28672m1 − 4096m2 − 86016)h30 + (22016m21 − 6656m1 m2 − 124416m1 + 512m22 + 18432m2 + 177664)h20 + (8128m31 − 3840m21 m2 − 67584m21 + 608m1 m22 + 21248m1 m2 + 185696m1 − 32m32 − 1696m22 − 28832m2 − 170016)h0 + (1300m41 − 864m31 m2 − 13872m31 + 216m21 m22 + 6904m21 m2 + 55224m21 − 24m1 m32 − 1152m1 m22 − 18264m1 m2 − 97152m1 + m42 + 64m32 + 1530m22 + 15904m2 + 64009) = 0

(9)

(192m21 − 80m1 m2 − 720m1 + 8m22 + 160m2 + 584)h1 − 1024h30 + (−1664m1 + 256m2 + 4608)h20 + (−784m21 + 224m1 m2 + 4512m1 − 16m22 − 640m2 − 6480)h0 + (−50m31 + 14m21 m2 + 522m21 − m1 m22 − 100m1 m2 − 1817m1 + 4m22 + 176m2 + 2116) = 0

(10)

(96m21 − 40m1 m2 − 360m1 + 4m22 + 80m2 + 292)h2 + 1024h30 + (1664m1 − 256m2 − 4608)h20 + (1168m21 − 384m1 m2 − 5952m1 + 32m22 + 960m2 + 7648)h0 + (242m31 − 118m21 m2 − 1986m21 + 19m1 m22 + 660m1 m2 + 5191m1 − m32 − 55m22 − 869m2 − 4379) = 0

(11)

4h3 − 8h0 + (−6m1 + m2 + 13) = 0

(12)

(96m21 − 40m1 m2 − 360m1 + 4m22 + 80m2 + 292)h4 − 1024h30 + (−1664m1 + 256m2 + 4608)h20 + (−1072m21 + 344m1 m2 + 5592m1 − 28m22 − 880m2 − 7356)h0 + (−242m31 + 118m21 m2 + 1890m21 − 19m1 m22 − 620m1 m2 − 4831m1 + m32 + 51m22 + 789m2 + 4087) = 0

(13)

(192m21 − 80m1 m2 − 720m1 + 8m22 + 160m2 + 584)h5 + 1024h30 + (1664m1 − 256m2 − 4608)h20 + (1168m21 − 384m1 m2 − 5952m1 + 32m22 + 960m2 + 7648)h0 + (338m31 − 182m21 m2 − 2418m21 + 33m1 m22 + 860m1 m2 + 5753m1 − 2m32 − 78m22 − 1002m2 − 4598) = 0.

(14)

Root Location Criteria Note that the first polynomial in the preceding system of polynomials is of fourth degree with coefficients that are functions of parameters m1 and m2 . Note that the system of equations is a triangular system. The solution to the fourth order polynomial will depend on the coefficients. So there are regions in the parameter space of m1 , m2 that yield real solutions. Use the root location criterion for the quartic equations (see Fuller[8]) a0 x4 + a1 x3 + a2 x2 + a3 x + a4 = 0 with coefficients all real and a0 > 0, based on the determinants of a ¯ ¯ a0 a1 ¯ ∆3 = ¯¯ 0 4a0 ¯ 4a0 3a1 ¯ ¯ ¯ ¯ ¯ ∆5 = ¯¯ ¯ ¯ ¯

a0 0 0 0 4a0

a1 a0 0 4a0 3a1

a2 a1 4a0 3a1 2a2

sequence of Sylvester matrices: ¯ a2 ¯¯ 3a1 ¯¯ 2a2 ¯ a3 a2 3a1 2a2 a3

a4 a3 2a2 a3 0

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

(15)

(16)

(17)

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Fig. 1. Root location criteria plot for length six filter coefficients case. The colored regions are defined by (a)∆3 > 0,(b)∆5 > 0,(c)∆7 > 0,respectively, and (d) is the region that satisfies ∆3 ,∆5 ,∆7 >0.

¯ ¯ ¯ ¯ ¯ ¯ ¯ ∆7 = ¯¯ ¯ ¯ ¯ ¯ ¯

a0 0 0 0 0 0 4a0

a1 a0 0 0 0 4a0 3a1

a2 a1 a0 0 4a0 3a1 2a2

a3 a2 a1 4a0 3a1 2a2 a3

a4 a3 a2 3a1 2a2 a3 0

0 a4 a3 2a2 a3 0 0

0 0 a4 a3 0 0 0

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯. ¯ ¯ ¯ ¯ ¯ ¯

(18)

The quartic equation (15) has all real roots if and only if anyone of the following conditions holds ∆3 > 0, ∆5 > 0, ∆7 > 0 ∆3 > 0, ∆5 > 0, ∆7 = 0 ∆3 > 0, ∆5 = 0, ∆7 = 0 ∆3 = 0, ∆5 = 0, ∆7 = 0.

(19)

Note that if the first condition is satisfied, then all the roots of the quartic are real and simple (multiplicity is one). Therefore, it becomes possible to find a region of parameter space, where each point yields four real solutions. Furthermore, the fourth degree equation will have only two real roots and a pair of complex roots if and only if any one of the following conditions holds, ∆7 < 0 ∆5 < 0, ∆7 = 0.

(20)

Substitute the coefficients of the parametrized quartic equation (9) into (16), (17) and (18) to find the determinants ∆3 , ∆5 and ∆7 that generate Figure 1 with parameters m1 and m2 as axes. In Figure 1(d), the green colored region gives four real solutions and in Figure 1(c) the white region corresponding to ∆7 < 0 gives two complex solutions. The zone with all real roots can be obtained by solving the inequalities ∆5 > 0 and ∆7 > 0, in this case. One of the boundaries for m2 is defined

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from 1 2 m , (21) 2 1 which is from ∆5 and the other, determined by ∆7 , is defined from q m2 < 5m1 − 4 − −m21 + 10m1 − 9. (22) √ √ √ √ The range of m1 extends from 5 − 15 to 5 + 15 and the corresponding range of m2 is from 20 + 5 15 to 20 − 5 15. The two complex roots zone also provides two real solutions and this region is described by q q (23) 5m1 − 4 − −m21 + 10m1 − 9 < m2 < 5m1 − 4 + −m21 + 10m1 − 9. m2 >

The valid intervals for m1 is [1, 9] and the corresponding m2 lies in the interval [1, 41]. Inequality (23) is not linear, and the maximum and minimum values of m2 are 41.3961 and 0.6039 respectively. From the above observation, it is interesting to note that the real solution boundary is generated by 1 2 m , 2 1 which is the relation between m2 and m1 when there are two vanishing moments. Therefore, when the parameters are closer to this boundary, the corresponding wavelets are smoother. m2 =

Special Case With parameters m1 = 1 and m2 = 1 one obtains the Haar wavelet and a translated one when parameters are m1 = 9 m2 = 41. With an additional vanishing moment, the sum rule yields −3h0 + 2h1 − h2 + h4 − 2h5 = 0.

(24)

Add (24) to the linear system (7) and solve with quadratic equations (8) again. The relation between m1 and m2 , 1 2 m , 2 1 generates the boundary of real roots region. For the special case of Daubechies wavelet, the third vanishing moment gives one more linear equation m2 =

−9h0 + 4h1 − h2 − h4 + 4h5 = 0.

(25)

Substituting the parameterized filter coefficients into equations (24) and (25), one gets two real solution sets for m1 and m2 m1 = 1.6348, m2 = 1.3363 m1 = 8.3652, m2 = 34.9883. The first solution generates the db3 plot in Figure 2 and the other one generates its translated version. C. Length Eight Case For the eight filter coefficient case there are two vanishing moments and thus the second discrete moment can be determined by the first discrete moment: 1 m2 = m21 . 2 The linear equations are h0 + h1 + h2 + h3 + h4 + h5 + h6 + h7 = 2 h0 − h1 + h2 − h3 + h4 − h5 + h6 − h7 = 0 4h0 − 3h1 + 2h2 − h3 + h5 − 2h6 + 3h7 = 0 h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 = m1 1 h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 = m21 2 h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 = m3

(26)

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Fig. 2. Special case of filter coefficients length six. The bold line is defined by the relation m2 = 1/2m21 and the black point in the left corner is the first parameter set (1.6348, 1.3363) that gives db3 and the translated version is the other point (8.3652, 34.9883) in the right corner.

and the quadratic equations (from orthogonality) are: h0 h2 + h1 h3 + h2 h4 + h3 h5 + h4 h6 + h5 h7 = 0 h0 h4 + h1 h5 + h2 h6 + h3 h7 = 0 h0 h6 + h1 h7 = 0

(27)

To parametrize eight filter coefficients, apply SINGULAR with lexicographical ordering and ranking h7 > ... > h1 > h0 to the sets of equations defined in (26) and (27) and obtain the reduced Gr¨obner basis as a triangular system of polynomial equations. The first polynomial with h0 as indeterminate and coefficients parametrized by m1 and m3 is: 1358954496h40 + (212336640m21 − 707788800m1 − 56623104m3 − 84934656)h30 + (20570112m41 − 177389568m31 − 9732096m21 m3 + 518750208m21 + 40108032m1 m3 − 641433600m1 + 1179648m23 − 19464192m3 + 368934912)h20 + (922752m61 − 12524544m51 − 663552m41 m3 + 67977216m41 + 6100992m31 m3 − 187077120m31 + 156672m21 m23 − 20173824m21 m3 + 266489856m21 − 718848m1 m23 + 29601792m1 m3 − 167380992m1 − 12288m33 + 681984m23 − 18247680m3 + 23639040)h0 + (26325m81 − 538488m71 − 23328m61 m3 + 4830624m61 + 357264m51 m3 − 24799104m51 + 7776m41 m23 − 2284416m41 m3 + 79502620m41 − 79488m31 m23 + 7778048m31 m3 − 162536640m31 − 1152m21 m33 + 307104m21 m23 − 14686848m21 m3 + 206735040m21 + 5888m1 m33 − 533376m1 m23 + 14114880m1 m3 − 150128640m1 + 64m43 − 7680m33 + 351360m23 − 4924800m3 + 48114000) = 0.

(28)

The triangular system is similar to that in the six coefficients case in (9) to (14). Since the other equations are too lengthy they are omitted here but can be found in Appendix. After solving (28) for h0 , substitute this value into the next equation and solve. Continue the process to generate the complete solution set of coefficients. Similar to the six coefficients case, there are also certain regions defined by the valid ranges of m1 and m3 . Using the root location criterion (19) and (20), get all real solutions and the set of two real accompanied with two complex solution regions shown in Figure 3. The subresultant matrices ∆5 , ∆7 are twelfth and twenty-fourth degree polynomials and there are no closed forms for real roots region.

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Fig. 3. Root location criteria plot for length eight filter coefficients case.The colored regions are defined by (a)∆3 > 0,(b)∆5 > 0,(c)∆7 > 0,respectively, and (d) is the region that satisfies ∆3 ,∆5 ,∆7 >0 (only the region pointed by arrow in the middle). Since these plots are numerical results, the regions are not very accurate when the determinants ∆5 , ∆7 are too large.

Special Case One more vanishing moment, generated by the sum rule 3, yields the linear equation −9h0 + 4h1 − h2 − h4 + 4h5 − 9h6 + 16h7 = 0. Applying SINGULAR with lexicographical ordering and ranking h7 > ... > h1 > h0 > m3 to the preceding equation and the sets of equations defined in (26) and (27), the first polynomial, in the Gr¨obner base, is a polynomial in the variable m3 , with its coefficients parametrized by m1 , as given next. (−256m21 + 3584m1 − 12800)m23 + (96m51 − 672m41 − 7936m31 + 81536m21 − 190208m1 + 102144)m3 + (−9m81 + 1488m61 − 4368m51 − 66400m41 + 443072m31 − 1059584m21 + 1180032m1 − 544320) = 0,

(29)

and this is plotted in Figure 4. When the point m1 , m3 is close to the real-valued solution set for the preceding equation, the associated wavelet will be smooth. For the special case of Daubechies wavelet, one more vanishing moment is needed. Again, applying the sum rule, the fourth vanishing moment V3 yields −27h0 + 8h1 − h2 + h4 − 8h5 − +27h6 − 64h7 = 0. The polynomial equation in m3 obtained by applying SINGULAR with lexicographical ordering and ranking h7 > ... > h1 > h0 > m1 > m3 on the preceding two equations and the sets of equations defined in (26) and (27) is m83 − 980m73 + 362376m63 − 68283068m53 + 6824666674m43 − 362290145364m33 + 11165951983816m23 − 191266139512700m3 + 94299960728785 = 0. The solution that yields db4 corresponds to m3 = 0.50784, m1 = 2.0108.

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Fig. 4. Special case of filter coefficients of length eight. The bold line is defined by the relation between m3 and m1 described by Eq.(29) and the black dot is the first parameter set (2.0108, 0.50784) that gives db4.

This solution is indicated by the black dot in Figure 4. D. Length Ten Case The maximum number of vanishing moments for a length ten filter is five. To acquire two parameters, one gives up the highest two vanishing moments. The linear equations are h0 + h1 + h2 + h3 + h4 + h5 + h6 + h7 + h8 + h9 h0 − h1 + h2 − h3 + h4 − h5 + h6 − h7 + h8 − h9 −5h0 + 4h1 − 3h2 + 2h3 − h4 + h6 − 2h7 + 3h8 − 4h9 −25h0 + 16h1 − 9h2 + 4h3 − h4 − h6 + 4h7 − 9h8 + 16h9 h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 + 8h8 + 9h9

=2 =0 =0 =0 = m1 1 h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 + 64h8 + 81h9 = m21 2 h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 + 512h8 + 729h9 = m3 ,

(30)

and the quadratic equations (from orthogonality) are h0 h2 + h1 h3 + h2 h4 + h3 h5 + h4 h6 + h5 h7 + h6 h8 + h7 h9 = 0 h0 h4 + h1 h5 + h2 h6 + h3 h7 + h4 h8 + h5 h9 = 0 h0 h6 + h1 h7 + h2 h8 + h3 h9 = 0 h0 h8 + h1 h9 = 0.

(31)

As in the length eight case, use reduced Gr¨obner bases with lexicographic order and ranking h9 > ...h1 > h0 using m1 and m3 as parameters. In the resulting triangular system of equations, the first equation only involves h0 as indeterminate with coefficients parametrized by m1 and m3 . After solving the first equation for h0 , substitute h0 into the next equation and solve. Continue the process to get the other coefficients. Since this triangular system of equations is too lengthy, it is omitted here but is listed in Appendix. The first equation in the triangular system is a quartic equation and therefore the root location criteria in (19) and (20) can

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Fig. 5. Root location criteria plot for length ten filter coefficients case. The colored regions are defined by (a)∆3 > 0,(b)∆5 > 0,(c)∆7 > 0, respectively, and (d) defines the region ∆3 ,∆5 ,∆7 >0 (only the region pointed by arrow in the middle). Since the criterion ∆7 > 0 is much more constrained than the others, the region that contains four real roots is determined by ∆7 .

be used. Since the coefficients of the triangular system become much more complex when the filter length is greater than 10, the numerical results are very sensitive to the change of parameters. In Figure 5 the edge should be smooth, for accuracy of computing. As in the length eight case, there are no closed form solutions for these roots location regions. Special Case If the fourth vanishing moment is added, the following linear equation is generated by the sum rule: −125h0 + 64h1 − 27h2 + 8h3 − h4 + h6 − 8h7 + 27h8 − 64h9 = 0.

(32)

Along with linear system (30) and quadratic equations (31), the above additional condition gives an equation in terms of m1 and m3 after applying SINGULAR with lexicographic order and ranking h9 > ...h1 > h0 > m3 on the system of equations above: 1024m43 + (−3072m31 + 55296m21 − 489472m1 + 1419264)m33 + (2880m61 − 93312m51 + 1540224m41 − 15303168m31 + 97677312m21 − 358511616m1 + 548785152)m23 + (−864m91 + 31104m81 − 496512m71 + 3768768m61 − 4056192m51 − 176214528m41 + 1512364544m31 − 11 10 5357366784m21 + 8252955648m1 − 4229148672)m3 + (81m12 1 − 2916m1 + 40716m1 −

155520m91 − 2354328m81 + 31658688m71 − 102669504m61 − 590398848m51 + 6210049216m41 − 22429995264m31 + 41210318592m21 − 39607335936m1 + 16394918400) = 0

(33)

The real-valued solutions to this equation is represented in parameter space by a bold line in Figure 6 . Note that this relation between m1 and m3 is along the boundary of the region defined by ∆7 > 0. When the parameters are closer to this line, the associated wavelets are smoother. Daubechies wavelet db5 has maximum number of vanishing moments: five. The linear

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8

9

Fig. 6. Special case of filter coefficients length ten. The bold line defines relation between m3 and m1 described in equation (29) and the black point is the parameter set (2.3878, 1.7018) that gives db5.

equation given by the fifth vanishing moment V4 is −625h0 + 256h1 − 81h2 + 16h3 − h4 − h6 + 16h7 − 81h8 + 256h9 = 0. Treating two parameters as variables to solve the linear equations, yields a sixteenth degree equation for m3 and an equation relating m1 in terms of m3 . These two equations give four real and twelve complex roots. The first real root m3 = 1.7018, m1 = 2.3878 gives Daubechies wavelet db5 which is indicated by a black dot in Figure 6. E. Length Twelve Case For a length twelve filter, there are four vanishing moments after the highest two are given up. By [1, Theorem 3.4], the fourth discrete moment m4 can be determined by the odd-indexed moments m1 and m3 , in this case, 3 m4 = − m41 +2m1 m3 . 8

11

The linear equations are h0 + h1 + h2 + h3 + h4 + h5 + h6 + h7 + h8 + h9 + h10 + h11 = 2 h0 − h1 + h2 − h3 + h4 − h5 + h6 − h7 + h8 − h9 + h10 − h11 = 0 −5h0 + 4h1 − 3h2 + 2h3 − h4 + h6 − 2h7 + 3h8 − 4h9 + 5h10 − 6h11 −25h0 + 16h1 − 9h2 + 4h3 − h4 − h6 + 4h7 − 9h8 + 16h9 − 25h10 + 36h11 −125h0 + 64h1 − 27h2 + 8h3 − h4 + h6 − 8h7 + 27h8 − 64h9 + 125h10 − 216h11 h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 + 8h8 + 9h9 + 10h10 + 11h11 h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 + 64h8 + 81h9 + 100h10 + 121h11 h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 + 512h8 + 729h9 + 1000h10 + 1331h11

=0 =0 =0 = m1 1 = m21 2 = m3

h1 + 16h2 + 81h3 + 256h4 + 625h5 + 1296h6 + 2401h7 + 4096h8 + 6561h9 + 10000h10 +14641h11 3 = − m41 +2m1 m3 8 (34)

and the quadratic equations (from orthogonality) are h0 h2 + h1 h3 + h2 h4 + h3 h5 + h4 h6 + h5 h7 + h6 h8 + h7 h9 + h8 h10 + h9 h11 = 0 h0 h4 + h1 h5 + h2 h6 + h3 h7 + h4 h8 + h5 h9 + h6 h10 + h7 h11 = 0 h0 h6 + h1 h7 + h2 h8 + h3 h9 + h4 h10 + h5 h11 = 0 h0 h8 + h1 h9 + h2 h10 + h3 h11 = 0 h0 h10 + h1 h11 = 0.

(35)

Applying SINGULAR with lexicographical ordering and ranking h11 > ... > h1 > h0 to the above sets of equations, the first polynomial equation in the triangular system of polynomial equations in the reduced Gr¨obner basis is an equation of fourth degree in h0 . As in the previous cases, the root location criteria in (19) and (20) can be used. The complex set of equations are omitted here. The zones ∆3 > 0, ∆5 > 0 and ∆7 > 0 are presented in Figures 7(a),(b),(c), respectively, and (d) defines the region ∆3 , ∆5 , ∆7 > 0. Special Case If the fifth vanishing moment V4 is added, the sum rule generates −625h0 + 256h1 − 81h2 + 16h3 − h4 − h6 + 16h7 − 81h8 + 256h9 − 625h10 + 1296h11 = 0.

(36)

Applying SINGULAR with lexicographical ordering and ranking h11 > ... > h1 > h0 > m3 to the preceding equation and the sets of equations defined in (34) and (35), the first polynomial, in the reduced Gr¨obner basis, is a polynomial in variable m3 , given by c0 m63 + c1 m53 + c2 m43 + c3 m33 + c4 m23 + c5 m3 + c6 = 0,

(37)

csk

where the coefficient are parametrized by m1 . The plot of the zero-set of Eq.(37) in the (m1 , m3 ) plane is shown in bold in Figure 8. When the parameters move closer to this curve, the associated wavelets are smoother. Applying the sum rule, the sixth vanishing moment V5 yields −3125h0 + 1024h1 − 243h2 + 32h3 − h4 + h6 − 32h7 + 243h8 − 1024h9 + 3125h10 − 7776h11 = 0

which, in addition to the set of equations 34, 35 and 36, can be used to solve for db6. Applying SINGULAR with lexicographical ordering and ranking h11 > ... > h1 > h0 > m1 > m3 , the first polynomial in the Gr¨obner basis is of 32nd degree in m3 . Solve this equation for m3 and substitute into the preceding relation (37). Then solve for m1 . It can then be seen that m3 = 3.382, m1 = 2.7643 leads to the Daubechies wavelet, db6, indicated as a black dot in Figure8. III. PARAMETRIZATION OF B IORTHOGONAL WAVELETS In the previous section, the parametrization is on FIR filters of the compactly supported orthonormal wavelets. The nonlinearity of phase response of those filters is undesirable in many applications. However, FIR filters that generate orthonormal wavelets cannot have linear phase except in the case of Haar wavelets (see [9, page.253-255] ). Therefore, parametrization of linear phase FIR filters capable of generating biorthogonal wavelets is considered. For a start, either one or two degrees of freedom with one or two discrete moments, are chosen for the parametrization. A brief review of biorthogonal wavelets and properties that will be needed for parametrization is given below. For details on biorthogonal wavelets, refer to [9],[10] and [11].

120

120

100

100

80

80

m3

m3

12

60

60

40

40

20

20

0

0

1

2

3

4

5

6

7

1

2

3

4

m1

5

6

7

m1

(a) ∆3 > 0

(b) ∆5 > 0

120

120

100

100

80

m3

m3

80

60

60 40 40 20 20 0 0 1

2

3

4

5

6

7

3

3.5

4

4.5

m1

(c) ∆7 > 0

5

5.5 m1

6

6.5

7

7.5

(d) ∆3, ∆5, ∆7 > 0

Fig. 7. Root location criteria plot for length ten filter coefficients case. The green colored regions are for (a)∆3 > 0,(b)∆5 > 0,(c)∆7 > 0, respectively, and case (d) includes the region ∆3 ,∆5 ,∆7 >0 (only the subregion pointed by arrow in the middle.) The numerical computation becomes much more sensitive to the variation of parameters in comparison to lower length cases.

A. A review of Biorthogonal Wavelets ˜ j − k) and dual mother wavelet ψ˜jk = 2j/2 ψ(2 ˜ j − k) For biorthogonal wavelets, the dual scaling function φ˜jk = 2j/2 φ(2 ˜ ˜ which constitute the dual subspaces Vj and Wj are, respectively, orthogonal to the wavelet ψjl and the scaling function φjl , i.e. hφ˜jk , ψjl i = 0, hψ˜jk , φjl i = 0.

(38)

The scaling function φjk is orthogonal to its dual φ˜jl , i.e hφ˜jk , φjl i = δkl ,

(39)

hψ˜jk , ψj 0 l i = δjj 0 δkl .

(40)

and so is the wavelet function ψjl to its dual ψ˜j 0 l , i.e

The properties (38), (39) and (40) are also known as biorthogonality. A scaling function and a mother wavelet satisfy the dilation equation and the wavelet equation, respectively, i.e. φ(x) =

N X

hk φ(2x − k) and ψ(x) =

k=0

˜ φ(x) =

N X k=0

N X

gk φ(2x − k)

(41)

˜ g˜k φ(2x − k)

(42)

k=0

˜ k φ(2x ˜ ˜ h − k) and ψ(x) =

N X k=0

13

120

100

m3

80

60

40

20

0

1

2

3

4

5

6

7

m1

Fig. 8. Special case of filter coefficients length twelve. The bold line is the relation between m3 and m1 described by Eq. (29), and the black dot is the parameter set (2.7643, 3.382) that gives db6.

˜ k , and g˜k and hk , obtained through the condition of exact where N can be odd or even. The relationships between gk and h reconstruction (refer to [9, p.262-263] (8.3.1) and (8.3.2)), are given as ˜ N −k gk = (−1)k+1 h g˜k = (−1)

k+1

hN −k ,

(43) (44)

where N = 2n − 1 in the [9]. Therefore, wavelet functions can be written as ψ(x) =

N X

˜ N −k φ(2x − k) (−1)k+1 h

(45)

k=0

and ˜ ψ(x) =

N X

˜ (−1)k+1 hN −k φ(2x − k).

(46)

k=0

˜ Substitute φ(x) and φ(x) defined in (41) and (42) into orthogonality relation between φ˜ and φ, i.e, Z ˜ − l)dx = δ0,l φ(x)φ(x

(47)

to obtain N X

˜k = 2 hk h

(48)

˜ k−2l = 0. l = 1, ..., (N − 1)/2. hk h

(49)

k=0

and a homogeneous equation N X k=0

14

˜ i satisfy the normalization constraints:(refer to Eq.(1)) The filter coefficients hi and its dual h N X

hk = 2 and

k=0

N X

˜ k = 2. h

(50)

k=0

Next, reduced vanishing moments for the case of symmetric filter coefficients are introduced. For convenience and clarity, the even length and odd length cases are treated separately. Reduced Vanishing Moments

Even length (N Odd) The condition of first p vanishing moments of associated wavelet is Z xl ψ(x)dx = 0, l = 0, 1, ..., p − 1.

(51)

Substitute the wavelet function in (45) into this condition to obtain the sum rule N X

˜ k = 0, l = 0, ..., p − 1, (−1)k k l h

(52)

˜ k = 0, N = 2n − 1. (−1)n−k (n − k)l h

(53)

k=0

which is equivalent to (see [1]) 2n−1 X k=0

˜ k , the vanishing moments of ψ generate the sum rule of h ˜ k . Furthermore, Since the filter coefficients gk of ψ are related to h since the filter coefficients are symmetric (linear phase), that is, ˜k = h ˜ N −k , k = 0, 1, ..., (N − 1)/2, h

(54)

equation (53) can be rewritten as n−1 X

˜ k = 0. (−1)k [(n − k)l − (k + 1 − n)l ]h

(55)

k=0

Note that the left hand side of Eq.(55) is identically zero when l = 0, implying that the first sum rule holds trivially. It can be proved by mathematical induction that (55), for odd values of integer l, becomes n−1 X

˜ k = 0, l = 1, 3, 5, 7... (−1)k [(n − k)l+1 − (k + 1 − n)l+1 ]h

(56)

k=0

From Eqs.(55) and (56), for a linear phase FIR filter of even length, the lth sum rule is the same as the (l + 1)th sum rule provided l = 1, 3, 5... is odd. Odd length (N even) The length of FIR filters associated with biorthogonal wavelets do not have to be even because double shift orthogonality does not hold anymore, i.e, N X ˜kh ˜ k−2l 6= 0. h k=0

If N is even (length is odd), the sum rule obtained in equation (53) has to be modified as 2n X

˜ k = 0, N = 2n. (−1)n−k+1 (n − k + 1)l h

(57)

k=0

Using symmetry property ˜k = h ˜ N −k , k = 0, 1, ..., N/2, h

(58)

the sum rule (57) becomes n−1 X k=0

˜k − h ˜ n = 0, (−1)k+1 [(n − k + 1)l + (k + 1 − n)l ]h

(59)

15

By mathematical induction one can prove, when l = 0, 2, 4, 6..., that Eq. (59) reduces to n−1 X

˜k − h ˜ n = 0. (−1)k+1 [(n − k + 1)l+1 + (k + 1 − n)l+1 ]h

(60)

k=0

From Eqs.(59) and (60), for a linear phase FIR filter of odd length, the lth sum rule is the same as the (l + 1)th sum rule provided l = 0, 2, 4... is even. The preceding discussions lead to the notion of reduced vanishing moments for symmetric biorthogonal wavelets. Since the symmetry property of filter coefficients makes the lth vanishing moment produce the same sum rule as the (l + 1)th , define the number p˜ of reduced vanishing moments as follows. Definition 1: In the linear phase biorthogonal wavelet system ψ(x) =

N X

gk ψ(2x − k),

k=0

the wavelet function ψ has p˜ reduced vanishing moments V˜l if it satisfies ½ Z ∞ for l = 0, 1, ..., 2(˜ p − 1) V˜l = xl ψ(x)dx = 0, for l = 0, 1, ..., 2˜ p −1 −∞

when N is odd; when N is even.

B. Discrete and Continuous Moments Discrete Moments for Even Length

The discrete moment of the analysis filter bank scaling function is defined as mn = hk = hN −k , implying that the discrete moment beomes

PN k=0

hk k n , N is odd. By symmetry,

N −1 2

mn =

X

(k n + (N − k)n )hk .

(61)

k=0

In particular, m0 =

N X

N −1 2

hk = 2, i.e

k=0

m1 =

N X

X

hk = 1

(62)

k=0 N −1 2

hk k = N

k=0

X

hk = N

(63)

k=0

N −1 2

m2 =

X

(k 2 + (N − k)2 )hk .

(64)

k=0

Define qn = k n + (N − k)n . Then q0 = 2, q1 = N and q2 = k 2 + (N − k)2 etc.. Solve for k in terms of q2 and substitute into qn , n = 3, 4, ..., 7 to get, 1 q3 = (3N q2 − N 3 ) 2 1 q4 = (q22 + 2N 2 q2 − N 4 ) 2 1 q5 = (5N q22 − N 5 ) 4 1 q6 = (q23 + 6N 2 q22 − 3N 4 q2 ) 4 1 q7 = (7N q23 + 7N 3 q22 − 7N 7 q2 + N 7 ) 8

(65) (66) (67) (68) (69)

From (66) and (68) one can write q22 =2q4 − 2N 2 q2 + N 4 ,

(70)

q32

(71)

=4q6 −

6N 2 q22

4

+ 3N q2 .

16

Substitute q22 and q23 into (67) and (69) to represent q3 , q5 and q7 as 1 q3 = (3N q2 − N 3 ) 2 1 q5 = (10N q4 − 10N 3 q2 + 4N 5 ) 4 1 q7 = (28N q6 − 70N 3 q4 + 84N 5 q2 − 34N 7 ) 8 From (62) and (63), m0 = 2, m1 = N. Substitute q3 , q5 and q7 into (61) to get the odd-indexed moments in terms of the smaller even-indexed moments, noting that m1 = N is a constant. 1 m3 = (3m1 m2 − m31 ) (72) 2 1 (73) m5 = (10m1 m4 − 10m31 m2 + 4m51 ) 4 1 m7 = (28m1 m6 − 70m31 m4 + 84m51 m2 − 34m71 ). (74) 8 Discrete Moments for Odd Length

For the odd length filter coefficient case, Eqs. (72), (73) and (74) still hold. With symmetry hk = hN −k , in the N even case, here, the discrete moments are rewritten as mn =

N X

N 2

n

hk k =

k=0

−1 X

(k n + (N − k)n )hk + (

k=0

N n ) hN/2 , N is even. 2

n

As in the even length case, let qn = k + (N − k)n and obtain the same q3 , q4 , ..., q7 as described above. Substitute q3 = 1 3 2 (3N q2 − N ) into N 2

m3 =

−1 X

(k 3 + (N − k)3 )hk + (

k=0

N 3 ) hN/2 , 2

to get 3m1 m2 − m31 2 The expressions for m5 , m7 in (73) and (74) are also verified. The odd-indexed discrete moments are linearly dependent on the even-indexed moments in both the even and odd length cases. Next the relationship between continuous and discrete moments of the scaling function and its dual is investigated. m3 =

The Relationship Between Discrete and Continuous Moments for Biorthogonal wavelet ˜ The continuous and discrete moments of the dual scaling function φ(x) are defined, respectively, by Z ˜ ˜ n = xn φ(x)dx M X ˜ k kn . m ˜n = h

(75) (76)

k

They also satisfy the recursive equation obtained by Strang & Nguyen[2, p.396] and [1]: ¶ n µ X 1 n ˜n = ˜ n−i M m ˜ iM i 2n+1 − 2 i=1 n−1 Xµ n ¶ n+1 ˜ ˜ n−i . m ˜ n = (2 − 2)Mn − m ˜ iM i

(77) (78)

i=1

From the proved results in [1], if p moments of the wavelet ψ vanish, then X ˜ − k) = M ˜ n , f or 0 ≤ n ≤ p − 1. (x − k)n φ(x k

(79)

17

and X k

¶ n µ X n ˜ i , f or 0 ≤ n ≤ p − 1. k φ(x − k) = (−1)i xn−i M i n˜

(80)

i=0

˜ then φ˜ and M ˜ n in the above equations (79) and (80) will be replaced If there are p vanishing moments for the dual wavelet ψ, by φ and Mn . For the proof, see Sweldens &Piessens[12]. With the above knowledge a relationship between dual continuous moments of biorthogonal scaling functions can be obtained as the following. ˜ − k) >k∈Z = 0. Let p ∈ N Theorem 1: Let φ, φ˜ ∈ L2 (R) be a scaling functions and its dual with biorthogonality < φ(x)φ(x be odd, and the associated wavelet function ψ has p + 1 vanishing moments. Then µ ¶ p X p ˜ p+1 + Mp+1 = ˜ p−i+1 + M ˜ i Mp−i+1 ). M (−1)i+1 (Mi M (81) i i=1

Proof: Inspired by [1], define ˜ − k) > for k ∈ Z. sk ,< x, φ(x)φ(x

(82)

˜ + k) >=< x − k, φ(x − k)φ(x) ˜ ˜ s−k =< x, φ(x)φ(x >=< x, φ(x − k)φ(x) >= s˜k .

(83)

By biorthogonality, s−k can be written as

Furthermore, since p is odd ∞ X

= =

k sk +

k=−∞ ∞ X p

k p s−k

k=−∞

k sk +

k=0 ∞ X

∞ X

p

k p sk −

k=0

0 X

k p sk +

∞ X

k=−∞ ∞ X p

∞ X

k=0

k=0

k p s−k +

k=0

k s−k +

0 X

k p s−k

k=−∞

k p s−k −

∞ X

k p sk

k=0

= 0.

(84)

Using (82), (80) and the definition of continuous moment, write ∞ X

k p sk =< x, φ(x)

k=−∞

∞ X

˜ − k) >= k p φ(x

k=−∞

¶ ¶ p µ p µ X X p p ˜ i Mp−i+1 . ˜ i < x, φ(x)xp−i >= (−1)i M (−1)i M i i

(85)

i=0

i=0

Furthermore, since s−k can be written as the dual of sk , derived in (83). Therefore, ¶ p µ ∞ ∞ X X X p ˜ p−i+1 . (−1)i Mi M k p s−k = k p s˜k = i k=−∞

(86)

i=0

k=−∞

Finally, from (84) along with the above results (85) and (86), ∞ X k=−∞ p µ X i=0

k p sk + p i

¶

∞ X

k p s−k =

k=−∞

˜ p−i+1 + M ˜ i Mp−i+1 ) = 0. (−1)i (Mi M

(87)

˜ 0 = 1, and when p = 1, 3, 5, there are p + 1 = 2, 4, 6 vanishing moments in the wavelet function ψ. For example, set M0 = M ˜ 2 + M2 =2M1 M ˜1 M ˜ 4 + M4 =4(M1 M ˜3 + M ˜ 1 M3 ) − 6M2 M ˜2 M ˜ 6 + M6 =6(M1 M ˜5 + M ˜ 1 M5 ) − 15(M2 M ˜4 + M ˜ 2 M4 ) + 20M3 M ˜3 M

(88) (89) (90)

Special Case: Linear Phase Substitute the recursive relationship between continuous and discrete moments, i.e (77) and (78), along with the relationship

18

between odd-indexed and even-indexed discrete moments for symmetric scaling function, i.e (72) and (73) into the preceding examples to get m ˜ 2 + m2 =m21

(91)

m4 =m41

m ˜4 + − 3m21 m2 + 2m ˜ 6 + 2m6 =2m61 − 15m41 m2

3m22 +

(92)

45m21 m22

−

45m32

−

15m21 m4

+ 30m2 m4

(93)

Orthogonal Case ˜ then Eq.(83) will become sk = s−k and M ˜ n = Mn . Equations (88) to (90) becomes If φ = φ, M2 =M12

(94)

M4 = − 3M22 + 4M1 M3 M6 =10M32 + 6M1 M5 −

(95) 15M2 M4 .

(96)

Substitute M2 in the above Eq.(94) into the other two equations to obtain M4 = − 3M14 + 4M1 M3 M6 =10M32

+ 6M1 M5 −

(97) 60M13 M3

+

45M16 ,

(98)

which is the orthogonal case in [1]. C. Parametrization of Biorthogonal wavelets Since the FIR filter length of biorthogonal wavelet does not need to be even, therefore both even and odd length cases will be discussed. Furthermore, the filter of dual scaling function φ˜ does not have to be the same length with φ. Vetterli & Herley[13] showed that in two channel, perfect reconstruction filter bank, where all filters have linear phase, the two analysis filters (φ and ψ) have one of following forms: (a) Both filters are symmetric and of odd lengths, differing by an odd multiple of 2. (b) One filter is symmetric and the other is antisymmetric; both lengths are even, and are equal or differ by an even multiple of 2. (c) One filter is of odd length, the other one of even length; both have all zeros on the unit circle. Either both filters are symmetric, or one is symmetric and the other one is antisymmetric. For more detail, refer to [11] and [14]. The wavelet function ψ and ψ˜ can be determined once one has scaling function φ and ˜ since the filter coefficients gk of wavelet function ψ can be determined by h ˜ k and coefficients g˜k of dual wavelet function φ, ˜ ψ can be determined by hk : ˜ N −k and g˜k = (−1)k+1 hN −k , gk = (−1)k+1 h

(99)

˜ k is thus the same with gk which is one of the above cases. where support k is [0, N ]. The length condition of h ˜ In the The parametrization of wavelets will become the parametrization of scaling function φ and its dual function φ. ˜ will be discussed. One will be even/even and following, parametrization by different length types of φ and φ˜ (noted as φ/φ) the other will be odd/odd case. Distribution of Reduced Vanishing Moments in the Symmetric Biorthogonal Case Since the analysis and synthesis filters for biorthogonal wavelet transform are different, there are multiple choices in the allocation of reduced vanishing moments. For a maximally flat FIR filter that is associated with a symmetric biorthogonal wavelet, the number of reduced vanishing moments is p˜ = L/2, when L = N + 1 is the filter length for odd N . For odd length case, p˜ = (L − 1)/2. The first reduced vanishing moment V˜0 is a requirement for the resolution of identity in the theory of wavelets and cannot be eliminated. However, the symmetric coefficients will make the first sum rule redundant for length even case. Therefore, in the following length even case parametrization, the first sum rule is omitted. (Note that for length odd case, the first sum rule is still useful.) The minimum number of reduced vanishing moments for both φ and φ˜ is 1 for even as well as odd length cases. Therefore, the number of degrees of freedom available for allocation is p˜− 1 for length even case and p˜− 2 for length odd case. For a length six symmetric biorthogonal wavelet, the maximum number of reduced vanishing moments is 6/2 = 3. The number of degrees of freedom to allocate them is p˜ − 1 = 2 implying that (1,3) and (2,2) are the two possible distributions of reduced ˜ For a length five case, the maximum number of reduced vanishing moments is (5 − 1)/2 = 2. vanishing moments in (φ, φ). Here the number of degrees of freedom to allocate them is p˜ − 2 = 0 which means there can only be (1,1) reduced vanishing ˜ moments in (φ, φ). Below, the highest two reduced vanishing moments, V˜p−1 and V˜p−2 ˜ ˜ , are given up to obtain either one or two degrees of freedom of parameters. For example, for a length eight case parametrization with one parameter, p˜ = 3. The number of degrees ˜ of freedom in their allocation is p˜ − 1 = 2. Therefore, the reduced vanishing moments could be either (1,3) or (2,2) in (φ, φ).

19

Setup The setup is similar to the case of parameterized orthogonal wavelets. The scaling functions in the biorthogonal system satisfy the normalization constraint (50): N X

hk = 2 and

N X

˜ k = 2. h

k=0

k=0

The orthogonality of integer translates of scaling function φ(x) andφ(x − l) does not hold in the biorthogonal case, and therefore the length of filter coefficients is not constrained to be even. The quadratic equation which comes from orthogonality ˜ is also not satisfied anymore. The biorthogonality between φ(x) and φ(x) reflected in Eqs.(48) and (49) are rewritten below, for convenience. N X

˜k = 2 hk h

(100)

˜ k−2l = 0. while l = 1, ..., (N − 1)/2. hk h

(101)

k=0

and N X k=0

Alongwith the sum rule in (53), a linear system that allows the parameterization of the filter coefficients is constructed. Length Even Case If the filter length for the scaling function is even, the symmetry constraint, hk = hN −k , halves the number coefficients. The normalization equation for φ becomes N −1 2

X

hk = 1,

(102)

˜ k = 1. h

(103)

(−1)n−k [(n − k)l − (k + 1 − n)l ]hk = 0,

(104)

k=0

and for φ˜ the normalization equation becomes N −1 2

X

k=0

Furthermore, the sum rule leads to n−1 X k=0

˜ where N = 2n − 1 and l = 1, ..., p − 1, p is the number of vanishing moments of ψ. 4/4 Case When the length of analysis filter coefficients of symmetric biorthogonal wavelet is four , one obtains one parameter by giving up the second reduced vanishing moment, V˜1 . The remaining reduced vanishing moment is V˜0 . The sum rule is eliminated. Therefore, the linear system obtained from Eqs.(102),(104) and discrete moment are h0 + h1 = 1 9h0 + 5h1 = m2 . Note that from Eqs.(62) and (63) m0 = 2, m1 = N . Solve this linear system to get m2 − 5 4 −m2 + 9 h1 = . 4 h0 =

(105)

20

˜ are from Eqs. (103) and (101): The linear and quadratic equations associated with dual filter h ˜0 + h ˜1 = 1 h ˜ 0 + h1 h ˜1 = 1 h0 h ˜ 1 + h1 h ˜ 0 = 0. h0 h

(106)

Substitute the result of (105) into above dual filter equations to obtain m2 − 5 2(m2 − 7) ˜ 1 = m2 − 9 . h 2(m2 − 7)

˜0 = h

(107)

One more reduced vanishing moment from Eq. (51) with l = 1 will yield the sum rule 3h0 − h1 = 0. This additional condition gives an unique parameter value, m2 = 6 and h0 = 0.25, h1 = 0.75, h2 = 0.75, h3 = 0.25 ˜ 0 = −0.5, h ˜ 1 = 1.5, h ˜ 2 = 1.5, h ˜ 3 = −0.5, h which is biorthogonal spline wavelet bio3.1 in [9, p.277]. D. 6/6 Case In the length six case, the maximum number of reduced vanishing moments is three. By giving up the highest, V˜2 , one degree of freedom for parametrization is obtained and the remaining number p˜ = 2. Since the minimum number of reduced vanishing moments for each of the scaling function filters is one, the only way to distribute the reduced vanishing moments is 1 in φ and 2 in φ˜ . Furthermore, if the highest two reduced vanishing moments are given up to obtain two degrees of freedom, then there is only one reduced vanishing moments left for the analysis filter or the synthesis filter. In the following, these conditions are discussed separately. Reduced Vanishing moments(1,2)

The two reduced vanishing moments in the dual wavelet function ψ˜ gives one sum rule for hk . The linear system for parametrization of analysis filter coefficients becomes h0 + h1 + h2 = 1 −5h0 + 3h1 − h2 = 0 25h0 + 17h1 + 13h2 = m2 . This system is solved for the h0k s : m2 − 14 16 m2 − 10 h1 = 16 −m2 + 20 h2 = . 8 h0 =

(108)

˜ the linear and quadrature equations are For the dual filter coefficients h, ˜0 + h ˜1 + h ˜2 = 1 h ˜ 0 + h1 h ˜ 1 + h2 h ˜2 = 1 h0 h ˜ 0 + h2 h ˜ 1 + (h0 + h1 )h ˜ 2 = 0. h2 h

(109)

21

˜k : Substitute (108) into the above system to solve for h 2 ˜ 0 = −m2 + 26m2 − 168 h 8(m2 − 16) 2 ˜ 1 = m2 − 22m2 + 120 h 8(m2 − 16) m 2 − 20 ˜2 = h . 2(m2 − 16)

As a special case, add one reduced vanishing moment in dual wavelet function ψ˜ to obtain a linear independent sum rule with l = 3, −35h0 + 9h1 − h2 = 0. Substitute equation (108) into above sum rule to get m2 = 15 and 1 5 , , 16 16 ˜ k = [ 3 , − 15 , h 8 8 hk = [

5 , 8 5 , 2

5 5 1 , , ] 8 16 16 5 15 3 , − , ], 2 8 8

˜ 0 and h ˜ 1 in the which corresponds to the maximum number of reduced vanishing moments case. Since the numerator of h parametrization have a common root m2 = 12, therefore 1 hk = [− , 8 ˜ k = [0, 0, h

1 1 1 , 1, 1, , − ] 8 8 8 1, 1, 0, 0],

which is the biorthogonal spline wavelet bio1.3. It is noted that the difference in length (even case) is either zero or an even multiple of 2. Reduced Vanishing moments(1,1), Two parameters

For the two-parameter case, both filters have only one vanishing moment left which is the minimum. Since the third discrete moment m3 is dependent on m2 (see 72), the fourth discrete moment m4 in the analysis filter or m ˜ 2 in the synthesis filter can be used as the second parameter. Using m2 and m4

Since the first sum rule is eliminated by the symmetry property, the linear equations are h0 + h1 + h2 25h0 + 17h1 + 13h2 625h0 + 257h1 + 97h2 ˜0 + h ˜1 + h ˜2 h

=1 = m2 = m4 = 1,

(110)

and quadratic equations are ˜ 0 + h1 h ˜ 1 + h2 h ˜2 = 1 h0 h ˜ 0 + h2 h ˜ 1 + (h0 + h1 )h ˜2 = 0 h2 h ˜ ˜ 1 = 0. h1 h0 + h0 h

(111)

In this system, there are six unknowns with four linear and three quadratic equations. Applying SINGULAR with lexicographical ˜2 > h ˜1 > h ˜ 0 > h2 > h1 > h0 to the above sets of equations, the coefficients, in terms of m2 and m4 , ordering and ranking h

22

obtained from the reduced Gr¨obner basis are h0 = −

h1 =

44m2 − m4 − 475 16

h2 = −

˜0 = h

40m2 − m4 − 423 48

46m2 − m4 − 525 24

1840m22 − 86m2 m4 − 39498m2 + m24 + 924m4 + 211923 15824m22 − 712m2 m4 − 346488m2 + 8m24 + 7800m4 + 1896048

2 2 ˜ 1 = 6072m2 − 270m2 m4 − 131682m2 + 3m4 + 2928m4 + 713925 h 2 15824m2 − 712m2 m4 − 346488m2 + 8m24 + 7800m4 + 1896048

˜ 2 = 46m2 − m4 − 525 . h 92m2 − 2m4 − 1026

(112)

Using m2 and m ˜2

Since there is only one reduced vanishing moment in both filters, the condition in Theorem 1 is not satisfied. Therefore, the discrete moment m ˜ 2 is still independent of m2 . The linear equations of analysis and synthesis filter coefficients are h0 + h1 + h2 = 1 25h0 + 17h1 + 13h2 = m2 ˜0 + h ˜1 + h ˜2 = 1 h ˜ 0 + 17h ˜ 1 + 13h ˜2 = m 25h ˜ 2,

(113)

˜ 0 + h1 h ˜ 1 + h2 h ˜2 = 1 h0 h ˜ 0 + h2 h ˜ 1 + (h0 + h1 )h ˜2 = 0 h2 h ˜ 0 + h0 h ˜ 1 = 0. h1 h

(114)

and the quadratic equations are

˜2 > h ˜1 > h ˜ 0 > h2 > h1 > h0 to the above sets of Applying SINGULAR with lexicographical ordering and ranking h equations, the first polynomial equation , parametrized by m2 and m ˜ 2 , obtained from the reduced Gr¨obner basis is a quadratic equation in h0 : (128m ˜ 2 − 2048)h20 + (−24m2 m ˜ 2 + 376m2 + 344m ˜ 2 − 5304)h0 + (m22 m ˜ 2 − 15m22 − 28m2 m ˜ 2 + 416m2 + 195m ˜ 2 − 2873) = 0.

Solve the above equation for h0 and substitute into the following equations to get the other coefficients. 4h1 + 12h0 + (−m2 + 13) = 0 4h2 − 8h0 + (m2 − 17) = 0 ˜ 0 + (16m (16m2 − 256)h ˜ 2 − 256)h0 + (−3m2 m ˜ 2 + 45m2 + 45m ˜ 2 − 663) = 0 ˜ 1 + (−48m (16m2 − 256)h ˜ 2 + 768)h0 + (5m2 m ˜ 2 − 83m2 − 71m ˜ 2 + 1157) = 0 ˜ 2 + (16m (8m2 − 128)h ˜ 2 − 256)h0 + (−m2 m ˜ 2 + 11m2 + 13m ˜ 2 − 119) = 0.

(115)

E. 8/8 case In the length eight case, the maximum number of reduced vanishing moments is four. By giving up the highest one, V˜3 , the reduced vanishing moments of analysis filter and synthesis filter can be (1,3) or (2,2) with one parameter. If one gives up one more reduced vanishing moment V˜2 for two parameters, then the allocation of reduced vanishing moments will become (2,1).

23

Reduced Vanishing moments (1,3)

˜ the linear equations for the filter coefficients of φ are When there are three reduced vanishing moments in ψ, h0 + h1 + h2 + h3 = 1 7h0 − 5h1 + 3h2 − h3 = 0 91h0 − 35h1 + 9h2 − h3 = 0 49h0 + 37h1 + 29h2 + 25h3 = m2 .

(116)

After solving the preceding linear system, one has m2 − 27 64 3m2 − 77 h1 = 64 m2 − 7 h2 = 64 −5m2 + 175 h3 = . 64 h0 =

(117)

The equations for the dual filter coefficients φ˜ are ˜0 + h ˜1 + h ˜2 + h ˜ 3 =1 h ˜ 0 + h1 h ˜ 1 + h2 h ˜ 2 + h3 h ˜ 3 =1 h0 h ˜ ˜ ˜ ˜ 3 =0 h2 h0 + h3 h1 + (h0 + h3 )h2 + (h1 + h2 )h ˜ 0 + h2 h ˜ 1 + h1 h ˜ 2 + h0 h ˜ 3 =0. h3 h

(118)

where the first one is from normalization and the first sum rule is eliminated by invoking the symmetry constraint on coefficients. ˜ k coefficients : Substitute hk in (117) into the preceding system of equations to get the h 2 3 ˜ 0 = m2 − 78m2 + 2031m2 − 17658 h 32(m2 − 29) 2 3 ˜ 1 = −3m2 + 230m2 − 5889m2 + 50358 h 32(m2 − 29) 2 3 ˜ 2 = 3m2 − 230m2 + 5905m2 − 50694 h 32(m2 − 29) 2 3 + 78m −m 2 − 2015m2 + 17066 2 ˜3 = h . 32(m2 − 29)

(119)

Reduced Vanishing moments (2,2)

If the reduced vanishing moments are equally distributed in the analysis and synthesis filters as (2,2), the linear equations for analysis filter coefficients are h0 + h1 + h2 + h3 = 1 7h0 − 5h1 + 3h2 − h3 = 0 49h0 + 37h1 + 29h2 + 25h3 = m2 .

(120)

and the linear equations for synthesis filter coefficients are ˜0 + h ˜1 + h ˜2 + h ˜3 = 1 h ˜ 0 − 5h ˜ 1 + 3h ˜2 − h ˜ 3 = 0. 7h

(121)

˜ 3 > ... > h ˜ 0 > h3 > ... > h0 to the above sets of equations Applying SINGULAR with lexicographical ordering and ranking h and the following quadratic equations ˜ 0 + h1 h ˜ 1 + h2 h ˜ 2 + h3 h ˜3 = 1 h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h3 )h ˜ 2 + (h1 + h2 )h ˜3 = 0 h2 h ˜ 0 + h2 h ˜ 1 + h1 h ˜ 2 + h0 h ˜ 3 = 0, h3 h

(122)

24

the first polynomial equation for h0 with coefficients parametrized by m2 , obtained in the Gr¨onber basis, is 2048h20 + (−32m22 + 1440m2 − 16192)h0 + (m32 − 75m22 + 1878m2 − 15704) = 0.

(123)

Solve this equation and substitute the roots into the following system of equations to get the other coefficients. 16h1 + 16h0 + (−m2 + 26) = 0 16h2 + 48h0 + (−m2 + 22) = 0 8h3 − 24h0 + (m2 − 32) = 0 ˜ 0 + 64h0 + (−m2 + 49m2 − 604) = 0 64h 2 ˜ 1 − 64h0 + (m2 − 45m2 + 512) = 0 64h 2

˜ 2 − 192h0 + (3m2 − 143m2 + 1704) = 0 64h 2 ˜ 3 + 192h0 + (−3m2 + 139m2 − 1676) = 0. 64h 2

(124)

˜ 1 and h ˜ 2 will be zeros and For the special case, when m2 = 23 the h 3 9 7 45 45 7 9 3 , − , − , , , − , − , ] 32 32 32 32 32 32 32 32 ˜ k = [0, 0, 1 , 3 , 3 , 1 , 0, 0], h 4 4 4 4 which is the biorthogonal spline wavelet bio3.3. The length difference is again an even multiple by 2. hk = [

Reduced Vanishing moments (1,2) and two parameters

The parametrization of biorthogonal wavelets with two parameters requires one more degree of freedom which results by giving up one more reduced vanishing moment. The distribution of reduced vanishing moments will then be (1,2). By Theorem 1, m ˜ 2 will be linear dependent on m2 , that is m ˜ 2 + m2 = m21 = N 2 . The only realization of the parametrization is with parameters m2 and m4 . Using m2 and m4

The linear equations for hk are h0 + h1 + h2 + h3 7h0 − 5h1 + 3h2 − h3 49h0 + 37h1 + 29h2 + 25h3 24010 + 1297h1 + 641h2 + 337h3

=1 =0 = m2 = m4

(125)

˜ k is and one linear equation for synthesis filter coefficients h ˜0 + h ˜1 + h ˜2 + h ˜ 3 = 1. h

(126)

˜ 0 + h1 h ˜ 1 + h2 h ˜ 2 + h3 h ˜3 = 1 h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h3 )h ˜ 2 + (h1 + h2 )h ˜3 = 0 h2 h ˜ 0 + h2 h ˜ 1 + h1 h ˜ 2 + h0 h ˜ 3 = 0. h3 h

(127)

The quadratic equations are

˜ 3 > ... > h ˜ 0 > h3 > ... > h0 to the above sets of equations, Applying SINGULAR with lexicographical ordering and ranking h the coefficients obtained in the Gr¨obner basis, parametrized by m2 and m4 are 79m2 − m4 − 1641 192 91m2 − m4 − 1953 h1 = 192 83m2 − m4 − 1729 h2 = 64 87m2 − m4 − 1897 h3 = − 64 h0 = −

(128)

25

and 3 2 2 2 2 ˜ 0 = 6715m2 − 164m2 m4 − 430126m2 + m2 m4 + 7058m2 m4 + 9185073m2 − 22m4 − 75948m4 − 65387286 h 2 2 −223040m2 + 5344m2 m4 + 9422496m2 − 32m4 − 112896m4 − 99509472 3 2 2 2 2 ˜ 1 = 7735m2 − 176m2 m4 − 500794m2 + m2 m4 + 7634m2 m4 + 10811073m2 − 22m4 − 82812m4 − 77819238 h 2 2 −223040m2 + 5344m2 m4 + 9422496m2 − 32m4 − 112896m4 − 99509472 3 2 2 2 2 ˜ 2 = 7735m2 − 176m2 m4 − 389274m2 + m2 m4 + 4962m2 m4 + 6131313m2 − 6m4 + 26748m4 − 28724598 h 2 2 223040m2 − 5344m2 m4 − 9422496m2 + 32m4 + 112896m4 + 99509472 3 2 2 2 2 ˜ 3 = 6715m2 − 164m2 m4 − 318606m2 + m2 m4 + 4386m2 m4 + 4442337m2 − 6m4 − 19116m4 − 14972454 . h 2 2 223040m2 − 5344m2 m4 − 9422496m2 + 32m4 + 112896m4 + 99509472

(129)

F. 10/10 case When the length l = 10, the maximum number of reduced vanishing moments is five. Give up the highest one to get one parameter m2 . The distribution of reduced vanishing moments can either be (1,4) or (2,3). If one gives up one more reduced vanishing moment to obtain two parameters m2 and m4 , the reduced vanishing moments can be distributed as (1,3) or (2,2). Note, m ˜ 2 will be dependent on m2 and cannot be a parameter. In the following, these cases are discussed separately. Reduced Vanishing moments (1,4)

Obtained from normalization equation, discrete moments and sum rules along with symmetry property, the linear equations are as follows, h0 + h1 + h2 + h3 + h4 −9h0 + 7h1 − 5h2 + 3h3 − h4 −189h0 + 91h1 − 35h2 + 9h3 − h4 −4149h0 + 1267h1 − 275h2 + 33h3 − h4 81h0 + 65h1 + 53h2 + 45h3 + 41h4

=1 =0 =0 =0 = m2 .

(130)

From the above, directly get the hk coefficients. h0 = h1 = h2 = h3 = h4 =

m2 − 44 256 5m2 − 216 256 2m2 − 81 64 21 64 −7m2 + 378 128

(131)

˜k Substitute the above hk into following linear and quadratic equations for h ˜0 + h ˜1 + h ˜2 + h ˜3 + h ˜4 = 1 h ˜ 0 + h1 h ˜ 1 + h2 h ˜ 2 + h3 h ˜ 3 + h4 h ˜4 = 1 h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h4 )h ˜ 2 + (h1 + h4 )h ˜ 3 + (h2 + h3 )h ˜4 = 0 h2 h ˜ ˜ ˜ ˜ ˜4 = 0 h4 h0 + h4 h1 + h3 h2 + h2 h3 + (h0 + h1 )h ˜ 0 + h2 h ˜ 1 + h1 h ˜ 2 + h0 h ˜3 = 0 h3 h ˜ 0 + h0 h ˜ 1 = 0, h1 h

(132)

26

˜ k coefficients below. to obtain the h 4 3 2 ˜ 0 = −m2 + 172m2 − 11098m2 + 318364m2 − 3425840 h 128(m2 − 46) 4 3 2 ˜ 1 = 5m2 − 856m2 + 54978m2 − 1569956m2 + 16817760 h 128(m2 − 46) 4 3 2 −5m + 854m − 54730m 2 2 2 + 1559696m2 − 16676280 ˜2 = h 64(m2 − 46) 4 3 2 5m − 854m + 54730m 2 2 2 − 1559664m2 + 16675320 ˜3 = h 64(m2 − 46) 4 3 2 −m + 171m − 10970m 2 2 2 + 312914m2 − 3348972) ˜4 = h 32(m2 − 46)

(133)

Reduced Vanishing Moments (2,3)

˜ the linear system When two reduced vanishing moments are allocated to wavelet function ψ and three to its dual ψ, h0 + h1 + h2 + h3 + h4 −9h0 + 7h1 − 5h2 + 3h3 − h4 −189h0 + 91h1 − 35h2 + 9h3 − h4 81h0 + 65h1 + 53h2 + 45h3 + 41h4 ˜0 + h ˜1 + h ˜2 + h ˜3 + h ˜4 h

=1 =0 =0 = m2

=1 ˜ ˜ ˜ ˜ ˜ −9h0 + 7h1 − 5h2 + 3h3 − h4 = 0,

˜ k . The quadratic equations are the same as in the previous contains second and third sum rules for hk and second sum rule for h ˜ 4 > ... > h ˜ 0 > h4 > ... > h0 to the above sets case (132). Applying SINGULAR with lexicographical ordering and ranking h of equations and (132), the first polynomial equation for h0 with coefficients parametrized by m2 , obtained in the Gr¨obner basis, is (−65536m2 + 2621440)h20 + (128m32 − 15232m22 + 603264m2 − 7948160)h0 + (−m42 + 168m32 − 10588m22 + 296680m2 − 3118403) = 0.

(134)

Solve the above equation for h0 and then substitute into the following equations obtained from the same Gr¨obner basis to get the other coefficients. m2 − 43 64 3m2 − 125 = −4h0 + 64 m2 − 23 = −4h0 + 64 −5m2 + 255 = 6h0 + 64 m3 − 125m22 + 5213m2 − 72521 = 2(m2 − 41)h0 − 2 256 3m32 − 371m22 + 15311m2 − 210815 = −2(3m2 − 121)h0 + 256 m22 − 78m2 + 1531 = 4h0 − 64 2m32 − 247m22 + 10184m2 − 140153 = 4(4m2 − 161)h0 − 64 3m32 − 369m22 + 15163m2 − 207969 . = −12(m2 − 40)h0 + 128

h1 = h0 + h2 h3 h4 ˜0 h ˜1 h ˜2 h ˜3 h ˜4 h

(135)

27

Reduced Vanishing Moments (1,3) and two Parameters

The second parameter m4 is obtained by giving up the fourth reduced vanishing moment V˜3 . The linear system becomes h0 + h1 + h2 + h3 + h4 = 1 −9h0 + 7h1 − 5h2 + 3h3 − h4 = 0 −189h0 + 91h1 − 35h2 + 9h3 − h4 = 0 81h0 + 65h1 + 53h2 + 45h3 + 41h4 = m2 6561h0 + 4097h1 + 2417h2 + 1377h3 + 881h4 = m4 ˜0 + h ˜1 + h ˜2 + h ˜3 + h ˜ 4 = 1. h

(136)

The quadratic equations are still the same: ˜ 0 + h1 h ˜ 1 + h2 h ˜ 2 + h3 h ˜ 3 + h4 h ˜4 = 1 h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h4 )h ˜ 2 + (h1 + h4 )h ˜ 3 + (h2 + h3 )h ˜4 = 0 h2 h ˜ 0 + h4 h ˜ 1 + h3 h ˜ 2 + h2 h ˜ 3 + (h0 + h1 )h ˜4 = 0 h4 h ˜ 0 + h2 h ˜ 1 + h1 h ˜ 2 + h0 h ˜ 3 = 0. h3 h ˜ 4 > ... > h ˜ 0 > h4 > ... > h0 to the above sets of equations, Applying SINGULAR with lexicographical ordering and ranking h ˜ i obtained in the Gr¨obner basis, parametrized by m2 and m4 , are the coefficients hi and h 130m2 − m4 − 4458 768 118m2 − m4 − 3942 =− 768 139m2 − m4 − 4833 = 192 133m2 − m4 − 4527 = 192 140m2 − m4 − 4968 =− 128

h0 = − h1 h2 h3 h4 and

˜0+ (−6945792m22 + 103296m2 m4 + 481019904m2 − 384m24 − 3576960m4 − 8327937024)h (53040m42 − 798m32 m4 − 5660624m32 + 3m22 m24 + 39936m22 m4 + 209895696m22 + 153m2 m24

+ 52038m2 m4 − 2982370608m2 − m34 − 8694m24 − 16558488m4 + 10367524800) = 0

˜1+ (−6945792m22 + 103296m2 m4 + 481019904m2 − 384m24 − 3576960m4 − 8327937024)h (−48144m42 + 762m32 m4 + 5095472m32 − 3m22 m24 − 38664m22 m4 − 187432752m22 − 141m2 m24 + 10662m2 m4 + 2644256592m2 + m34 + 8178m24 + 14372712m4 − 9167515200) = 0 ˜2+ (−3472896m22 + 51648m2 m4 + 240509952m2 − 192m24 − 1788480m4 − 4163968512)h (−60384m42 + 852m32 m4 + 6074240m32 − 3m22 m24 − 35388m22 m4 − 197030112m22 − 195m2 m24 − 614976m2 m4 + 1826316864m2 + m34 + 10380m24 + 28334736m4 + 7621879680) = 0 ˜3+ (−3472896m22 + 51648m2 m4 + 240509952m2 − 192m24 − 1788480m4 − 4163968512)h (60384m42 − 852m32 m4 − 6074240m32 + 3m22 m24 + 35388m22 m4 + 198766560m22 + 195m2 m24

+ 589152m2 m4 − 1945958976m2 − m34 − 10284m24 − 27445104m4 − 5561032320) = 0 ˜4+ (578816m22 − 8608m2 m4 − 40084992m2 + 32m24 + 298080m4 + 693994752)h (408m42 − 3m32 m4 − 47096m32 + 106m22 m4 + 1582504m22 + m2 m24 + 9529m2 m4 − 8031528m2 − 59m24 − 331956m4 − 250519392) = 0.

(137)

28

Reduced Vanishing moments (2,2) and two Parameters

The other way to allocate the reduced vanishing moments is by distributing equally between analysis and synthesis filters. The linear system becomes h0 + h1 + h2 + h3 + h4 −9h0 + 7h1 − 5h2 + 3h3 − h4 81h0 + 65h1 + 53h2 + 45h3 + 41h4 6561h0 + 4097h1 + 2417h2 + 1377h3 + 881h4

=1 =0 = m2 = m4

and ˜0 + h ˜1 + h ˜2 + h ˜3 + h ˜4 = 1 h ˜ 0 + 7h ˜ 1 − 5h ˜ 2 + 3h ˜3 − h ˜ 4 = 0. −9h ˜ 4 > ... > h ˜0 > The quadratic equations remain the same. Applying SINGULAR with lexicographical ordering and ranking h h4 > ... > h0 to the sets of equations, the first polynomial equation for h0 with coefficients parametrized by m2 and m4 , obtained in the Gr¨obner basis, is (−884736m22 + 31555584m2 + 294912m4 − 79626240)h20 + 2 + + 31863168m2 + 26496m2 m4 − 606450816m2 − 1152m24 − 4836096m4 + 1585090944)h0 + (−50673m42 + 780m32 m4 + 5335214m32 − 3m22 m24 − 38475m22 m4 − 193582242m22 − 150m2 m24 − 78588m2 m4 + 2628672426m2 + m34 + 8541m24 + 16484391m4 − 7572966453)

(−452736m32

3456m22 m4

=0

Solve the above equation for h0 and then substitute into the following equations to obtain the other coefficients. 192h1 + 576h0 + (127m2 − m4 − 4329) = 0 192h2 + (−139m2 + m4 + 4833) = 0 64h3 − 512h0 + (−131m2 + m4 + 4481) = 0 64h4 + 384h0 + (135m2 − m4 − 4713) = 0 ˜0+ (−34816m2 + 256m4 + 1209600)h (768m22 − 27392m2 − 256m4 + 69120)h0 + (399m32

−

3m22 m4

−

28409m22

− 23m2 m4 + 555903m2 + m24 + 4212m4 − 1749357) = 0

˜ 1 + (−6912m22 + 246528m2 + 2304m4 − 622080)h0 + (−104448m2 + 768m4 + 3628800)h (−1959m32 + 15m22 m4 + 135877m22 + 127m2 m4 − 2482695m2 − 5m24 − 21420m4 + 4347513) = 0 ˜ 2 + (3m22 − 104m2 − m4 + 135) = 0 192h ˜3+ (−8704m2 + 64m4 + 302400)h (1536m22

− 54784m2 − 512m4 + 138240)h0 +

(390m32 − 3m22 m4 − 27411m22 − 22m2 m4 + 523990m2 + m24 + 4130m4 − 1462239) = 0 ˜4+ (−17408m2 + 128m4 + 604800)h (−2304m22

+ 82176m2 + 768m4 − 207360)h0 +

(−381m32 + 3m22 m4 + 27501m22 + 13m2 m4 − 554901m2 − m24 − 3864m4 + 2044521) = 0.

Length Odd Case If the filter length for scaling function is odd, then the symmetry constraint hk = hN −k transforms the normalization equation for φ to N 2

2

−1 X

k=0

hk + hN/2 = 2,

(138)

29

For φ˜ the normalization equation becomes N 2

2

−1 X

˜k + h ˜ N/2 = 2. h

(139)

k=0

Furthermore, the equation from sum rule becomes n−1 X

˜k − h ˜ n = 0, (−1)n−k+1 [(n − k + 1)l + (k + 1 − n)l ]h

k=0

˜ where N = 2n and l = 0, 1, ..., p − 1, p is vanishing moments of ψ. With the symmetric coefficients, the discrete moments will become mn =

N X k=0

N 2

n

hk k =

−1 X

(k n + (N − k)n )hk + (

k=0

N n ) hN/2 , 2

where N is even and n is the order of discrete moment. The odd-indexed moments are linearly dependent on even-indexed moments, described in equation (72) to (74). Since m0 and m1 are constant, m2 will be chosen as parameter. In the following, when the length is five, the maximum number of reduced vanishing moments is p˜ = (5 − 1)/2 = 2 and the the number of degree of freedom to allocate reduced vanishing moments is p˜ − 2 = 0, which means that length 5/5 case has no degree of freedom for parametrization. G. 5/5 case ˜ is (1,1) for this case. The linear equations obtained from normalization, From above the only possible allocation in (ψ, ψ) sum rules and discrete moment are 2h0 + 2h1 + h2 −2h0 + 2h1 − h2 16h0 + 10h1 + 4h2 ˜ 0 + 2h ˜1 + h ˜2 2h

=2 =0 = m2

=2 ˜ ˜ ˜ −2h0 + 2h1 − h2 = 0.

(140)

˜ 0 + 2h1 h ˜ 1 + h2 h ˜2 = 2 2h0 h ˜ 2 + h1 h ˜ 1 + h2 h ˜ 0 = 0. h0 h

(141)

The quadratic equations are

Solve the above sets of equations to get m2 = 7 or m2 = 9. For m2 = 7, the corresponding filter coefficients are h = [−0.25 0.5 1.5 0.5 − 0.25] ˜ = [0 0.5 1 0.5 0]. h

(142)

When m2 = 9, the two filter coefficients are the same but just interchanged. This is exactly the case of biorthogonal spline wavelet bio 2.2 [9]. H. 7/7 case Reduced Vanishing Moment (1,1)

In this case, the maximum number of reduced vanishing moments will be three. Give up one reduced vanishing moment to obtain one degree of freedom for parametrization. The remaining number of reduced vanishing moments is two and the ˜ The linear number of degree of freedom for allocation is 0. Therefore, the only way to allocate the moment is (1,1) in (φ, φ). system becomes: 2h0 + 2h1 + 2h2 + h3 = 2 2h0 − 2h1 + 2h2 − h3 = 0 36h0 + 26h1 + 20h2 + 9h3 = m2 ˜ 0 + 2h ˜ 1 + 2h ˜2 + h ˜3 = 2 2h ˜ 0 − 2h ˜ 1 + 2h ˜2 − h ˜ 3 = 0. 2h

(143)

30

The quadratic equations are ˜ 0 + 2h1 h ˜ 1 + 2h2 h ˜ 2 + h3 h ˜3 = 2 2h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h2 )h ˜ 2 + h1 h ˜3 = 0 h2 h ˜ 2 + h1 h ˜ 1 + h2 h ˜ 0 = 0. h0 h

(144)

˜ 3 > ... > h ˜ 0 > h3 > ... > h0 to the above sets of equations, Applying SINGULAR with lexicographical ordering and ranking h the first polynomial equation for h0 with coefficients parametrized by m2 and the other polynomial equations obtained in the Gr¨obner basis are (−16m2 + 240)h20 + (m22 − 36m2 + 323)h0 = 0 8h1 + 16h0 + (−m2 + 19) = 0 2h2 + 2h0 − 1 = 0 4h3 − 16h0 + (m2 − 23) = 0 ˜ 0 + (16m2 − 240)h0 + (−m2 + 36m2 − 323) = 0 (−16m2 + 336)h 2 ˜ 1 + (−8m2 + 120)h0 + (m2 − 17) = 0 (−4m2 + 84)h ˜ 2 + (−16m2 + 240)h0 + (m2 − 28m2 + 155) = 0 (−16m2 + 336)h 2 ˜ 3 + (8m2 − 120)h0 + (m2 − 25) = 0. (−2m2 + 42)h

(145)

Use f acstd command in SINGULAR to factorize the above equations and get, in the first part, m2 − 19 1 −m2 + 23 , h2 = , h3 = 8 2 4 m22 − 36m2 + 323 ˜ m2 − 17 ˜ m22 − 28m2 + 155 ˜ m2 − 25 ˜ h0 = , h1 = , h2 = , h3 = . −16m2 + 336 4m2 − 84 16m2 − 336 2m2 − 42 h0 = 0, h1 =

(146)

and, in the second part, m2 − 19 m2 − 44m2 + 443 m2 − 11 m22 − 36m2 + 323 , h1 = , h2 = 2 , h3 = 16m2 − 240 4m2 − 60 −16m2 + 240 2m2 − 30 −m + 17 1 2 ˜ 0 = 0, h ˜1 = ˜2 = , h ˜ 3 = m2 − 13 . h , h (147) 8 2 4 Note that the preceding two cases correspond to the length 7/5 and 5/7 cases. For the special case, add one more reduced vanishing moment on ψ˜ to get one more sum rule for hk , that is h0 =

20h0 − 10h1 + 4h2 − h3 = 0.

(148)

Substitute the above parameterized coefficients into this sum rule to get m2 = 20 and 1 1 , h 2 = , h3 = 8 2 3 ˜ −3 ˜ 5 ˜ ˜ h0 = , h1 = , h2 = , h3 = 16 4 16

h0 = 0, h1 =

3 , 4 5 . 2

(149)

I. 9/9 case When the length is longer, there are more possibilities for allocating the reduced vanishing moments. The maximum number of reduced vanishing moments for length nine is four, and there is one parametrization with one parameter with allocation (1,2) since the number of degree of freedom for allocation is (˜ p − 2 = 1). If the parametrization is with two parameters, the number of degree freedom to allocate reduced vanishing moments is p˜ − 2 = 0. When the the number of reduced vanishing ˜ These different cases are discussed in moments is maximum, one has the special cases of (3,1) and (2,2) moments in (φ, φ). the following subsections.

31

Reduced vanishing moments (1,2)

When there are two reduced vanishing moments in ψ˜ there are two sum rules for hk . The linear equations are 2h0 + 2h1 + 2h2 + 2h3 + h4 = 2 2h0 − 2h1 + 2h2 − 2h3 + h4 = 0 −34h0 + 20h1 − 10h2 + 4h3 − h4 = 0 64h0 + 50h1 + 40h2 + 34h3 + 16h4 = m2 ˜ 0 + 2h ˜ 1 + 2h ˜ 2 + 2h ˜3 + h ˜4 = 2 2h ˜ 0 − 2h ˜ 1 + 2h ˜ 2 − 2h ˜3 + h ˜ 4 = 0. 2h

(150)

The quadratic equations are ˜ 0 + 2h1 h ˜ 1 + 2h2 h ˜ 2 + 2h3 h ˜ 3 + h4 h ˜4 = 2 2h0 h ˜ 0 + h3 h ˜ 1 + (h0 + h4 )h ˜ 2 + (h1 + h3 )h ˜ 3 + h2 h ˜4 = 0 h2 h ˜ 0 + h3 h ˜ 1 + h2 h ˜ 2 + h1 h ˜ 3 + h0 h ˜4 = 0 h4 h ˜ 0 + h1 h ˜ 1 + h0 h ˜ 2 = 0. h2 h

(151)

˜ 4 > ... > h ˜ 0 > h4 > ... > h0 to the above sets of equations Applying SINGULAR with lexicographical ordering and ranking h and factorizing the polynomial equations obtained in the Gr¨obner basis, the first set of coefficients is m2 − 34 m2 − 32 m2 − 50 m2 − 40 , h2 = , h3 = − , h4 = − 32 16 32 8 m32 − 99m22 + 3270m2 − 36040 ˜ −m32 + 97m22 − 3140m2 + 33920 ˜ m2 − 28 ˜ , h1 = , h2 = , h0 = 64m2 − 2304 32m2 − 1152 4m2 − 144 3 2 3 2 ˜ 4 = −m2 + 99m2 − 3254m2 + 35336 . ˜ 3 = m2 − 97m2 + 3156m2 − 34496 , h h 32m2 − 1152 32m2 − 1152

h0 = 0, h1 =

(152)

and the other set is 8192h20 + (−64m22 + 3904m2 − 59520)h0 + (m32 − 99m22 + 3270m2 − 36040) = 0 32h1 + (−m2 + 34) = 0 16h2 + 64h0 + (−m2 + 32) = 0 32h3 + (m2 − 50) = 0 8h4 − 48h0 + (m2 − 40) = 0 ˜0 = 0 h ˜ 1 + 128h0 + (−m22 + 65m2 − 1060) = 0 32h ˜ 2 − 128h0 + (m22 − 63m2 + 998) = 0 16h ˜ 3 − 128h0 + (m22 − 65m2 + 1044) = 0 32h ˜ 4 + 128h0 + (−m22 + 63m2 − 1006) = 0. 8h

(153)

For the second factorized parametrization, first solve the equation for h0 and them substitute into the other equations to get the coefficients. Reduced vanishing moments (1,1) and two Parameters

If one more reduced vanishing moments in analysis filter is given up, the fourth discrete moment m4 can be used as the second parameter. The linear equations are 2h0 + 2h1 + 2h2 + 2h3 + h4 = 2 2h0 − 2h1 + 2h2 − 2h3 + h4 = 0 64h0 + 50h1 + 40h2 + 34h3 + 16h4 = m2 4096h0 + 2402h1 + 1312h2 + 706h3 + 256h4 = m4 ˜ 0 + 2h ˜ 1 + 2h ˜ 2 + 2h ˜3 + h ˜4 = 2 2h ˜ 0 − 2h ˜ 1 + 2h ˜ 2 − 2h ˜3 + h ˜ 4 = 0. 2h

(154)

˜ 4 > ... > and the quadratic equations are still the same. Applying SINGULAR with lexicographical ordering and ranking h ˜ 0 > h4 > ... > h0 to the above sets of equations and factorizing the polynomial equations obtained in the Gr¨obner basis, the h

32

first set of coefficients is h0 = 0 100m2 − m4 − 2691 96 106m2 − m4 − 2889 h2 = 48 100m2 − m4 − 2643 h3 = 96 106m2 − m4 − 2913 h4 = − 24

h1 = −

3 2 2 2 2 ˜ 0 = − 10600m2 − 206m2 m4 − 882746m2 + m2 m4 + 11566m2 m4 + 24496425m2 − 29m4 − 162216m4 − 226520307 h 2 2 698752m2 − 13376m2 m4 − 38050752m2 + 64m4 + 364224m4 + 518002560 3 2 2 2 2 ˜ 1 = − 11236m2 − 212m2 m4 − 939584m2 + m2 m4 + 11938m2 m4 + 26184537m2 − 29m4 − 167958m4 − 243187353 h 2 2 349376m2 − 6688m2 m4 − 19025376m2 + 32m4 + 182112m4 + 259001280

˜ 2 = 106m2 − m4 − 2877 h 424m2 − 4m4 − 11604 3 2 2 2 2 ˜ 3 = 11236m2 − 212m2 m4 − 764896m2 + m2 m4 + 8594m2 m4 + 16671849m2 − 13m4 − 76902m4 − 113686713 h 2 2 349376m2 − 6688m2 m4 − 19025376m2 + 32m4 + 182112m4 + 259001280 3 2 2 2 2 ˜ 4 = 10600m2 − 206m2 m4 − 708058m2 + m2 m4 + 8222m2 m4 + 14944185m2 − 13m4 − 70776m4 − 95948307 h 2 2 349376m2 − 6688m2 m4 − 19025376m2 + 32m4 + 182112m4 + 259001280

(155)

The second set contains a quadratic equation in h0 with coefficients parametrized by m2 and m4 , i.e. (−73728m2 + 2064384)h20 + (−59904m22 + 576m2 m4 + 3327168m2 − 16320m4 − 46202688)h0 − (10600m32 − 206m22 m4 − 882746m22 + m2 m24 + 11566m2 m4 + 24496425m2 − 29m24 − 162216m4 − 226520307) = 0.

(156)

After solving this equation, substitute the solution for h0 into the equations below to get the other coefficients. 96h1 + 384h0 + (100m2 − m4 − 2691) = 0 48h2 − 192h0 + (−106m2 + m4 + 2889) = 0 96h3 − 384h0 + (−100m2 + m4 + 2643) = 0 24h4 + 240h0 + (106m2 − m4 − 2913) = 0 ˜0 = 0 h (3488m2 − 32m4 − 96192)˜h1 + (384m2 − 10752)h0 + (318m22 − 3m2 m4 − 17925m2 + 87m4 + 252531) = 0 ˜ 2 + (384m2 − 10752)h0 + (100m2 − m2 m4 − 5155m2 + 25m4 + 66159) = 0 (−1744m2 + 16m4 + 48096)h 2

˜ 3 + (−384m2 + 10752)h0 + (−318m2 + 3m2 m4 + 16181m2 − 71m4 − 204435) = 0 (3488m2 − 32m4 − 96192)h 2 ˜ 4 + (−384m2 + 10752)h0 + (−100m2 + m2 m4 + 6027m2 − 33m4 − 90207) = 0 (−872m2 + 8m4 + 24048)h 2

(157)

Special Case: Reduced vanishing moments (2,2), (1,3)

When the number of remaining reduced vanishing moments is four, there are two ways to distribute the moments. One is to equally distribute on both sides and the other is to put more on one side. The smoothness of wavelet function will depend on the number of vanishing moments allocated to it. In the equally distributed case, that is (2,2), one has the following linear equations 2h0 + 2h1 + 2h2 + 2h3 + h4 = 2 2h0 − 2h1 + 2h2 − 2h3 + h4 = 0 −34h0 + 20h1 − 10h2 + 4h3 − h4 = 0 ˜ 0 + 2h ˜ 1 + 2h ˜ 2 + 2h ˜3 + h ˜4 = 2 2h ˜ 0 − 2h ˜ 1 + 2h ˜ 2 − 2h ˜3 + h ˜4 = 0 2h ˜ 0 + 20h ˜ 1 − 10h ˜ 2 + 4h ˜3 − h ˜ 4 = 0, −34h

(158)

and the quadratic equations are still the same as in above. Solve this system to get the special case h0 = 0.0534, h1 = −0.0338, h2 = 0.1564, h3 = 0.5338, h4 = 1.2058 ˜ 0 = 0, h ˜ 1 = −0.0912, h ˜ 2 = −0.0576, h ˜ 3 = 0.5912, h ˜ 4 = 1.1150. h

(159)

33

Fig. 9. An example of a better choice for the application in [1]. The left-hand side figure is the wavelet found from the parametrization with two parameters of length eight orthogonal wavelets. m1 = 7.51, m3 = 97.6 give SE=4.1209 and ME=0.4825. The right-hand side figure shows the result of [1] and Daubechies wavelet db4.

˜ k . The linear which is biorthogonal spline wavelet bio4.4. [15] In the (1,3) case, there are three sum rules for hk and one for h equations become 2h0 + 2h1 + 2h2 + 2h3 + h4 = 2 2h0 − 2h1 + 2h2 − 2h3 + h4 = 0 −34h0 + 20h1 − 10h2 + 4h3 − h4 = 0 −706h0 + 272h1 − 82h2 + 17h3 − h4 = 0 ˜ 0 + 2h ˜ 1 + 2h ˜ 2 + 2h ˜3 + h ˜4 = 2 2h ˜ 0 − 2h ˜ 1 + 2h ˜ 2 − 2h ˜3 + h ˜ 4 = 0, 2h

(160)

which when solved alongwith the quadratic equations, provide two sets of answers: h0 = −0.0204, h1 = −0.0561, h2 = 0.0943, h3 = 0.5561, h4 = 0.8522 ˜ 0 = 0, ˜ 1 = 0.2015, h ˜ 2 = −0.5535, h ˜ 3 = 0.2985, h ˜ 4 = 2.1070, h h

(161)

h0 = 0, h1 = 0.0263, h2 = 0.1776, h3 = 0.4737, h4 = 0.6447 ˜ ˜ ˜ 2 = −1.4773, h ˜ 3 = −0.2084, h ˜ 4 = 4.1644. h0 = −0.1049, h1 = 0.7084, h

(162)

and

IV. E XAMPLES The choice of mother wavelet is application-dependent [16]. Parametrization with two parameters offers more choices than parametrization with one parameter. In [1], an example on signal compression is given within the setting of wavelet decomposition level four. The square error and maximum error of the ’optimal’ wavelets Regensburger proposed is better than Daubechies wavelet (db4). Here, the same input signal with the same setting are tested and one can see the wavelet proposed here is better than Regensburger’s wavelet. The new SE = 4.1209 and ME = 0.4825 are better than Regensburger’s wavelet (SE= 5.9633 and ME = 0.5471) and Daubechies’ wavelet (SE= 18.5174 and ME= 1.0148). V. C ONCLUSIONS Filter bank parametrization is useful in performance evaluation with a large number of different wavelets, in a convenient manner. Recently, Regensburger and Scherzer [1] used one discrete moment as parameter in the parametrization of FIR filter coefficients to generate orthogonal wavelets after establishing a bijective relationship between continuous and discrete moments of the scaling function through use of Bell polynomials. However, except trivially, orthogonal wavelets cannot yield linear

34

phase like symmetric biorthogonal wavelets. Here the scaling and wavelet filters and their duals in a two-channel filter bank can be used. This is more complicated because the problem of generating discrete scaling functions and mother wavelets which differ in length may be encountered. A characterization theorem for continuous moments of the scaling function and its dual was formulated and proved for any biorthogonal system. Regensburger’s result emerged as a special case of this general result given here as Theorem 1. The possibility of using Gr¨obner bases in the parametrization of filter bank coefficients for orthogonal wavelets was considered and shown to lead to better performance evaluation in a generic image compression example that was used by Regensburger and Scherzer [1] to show the superiority of his parametrization approach with one discrete moment over the classical Daubechies orthogonal system. It is shown that Gr¨obner bases are effective in formulating and solving a parametrized system of filter coefficients used for generating biorthogonal wavelets. The important special case of symmetric biorthogonal wavelets are discussed in detail for FIR filters of different lengths. The systematic generation of filters provide a mechanism for generating a wide variety of wavelets including many belonging to distinguished classes for selection of optimum application-dependent mother wavelets [16]. A PPENDIX Length Eight Orthogonal Case: The equations left in the triangular system mentioned in length eight case are: (13824m41 − 163968m31 − 5376m21 m3 + 725888m21 + 32768m1 m3 − 1381632m1 + 512m23 − 53760m3 + 898560)h1 − 2359296h30 + (−405504m21 + 1671168m1 + 98304m3 − 958464)h20 + (−14400m41 + 82176m31 + 7680m21 m3 + 32768m21 − 22528m1 m3 − 686592m1 − 1024m23 − 27648m3 + 707328)h0 + (117m61 − 2802m51 − 60m41 m3 + 24296m41 + 1184m31 m3 − 99320m31 + 8m21 m23 − 7184m21 m3 + 206960m21 − 128m1 m23 + 14912m1 m3 − 217320m1 + 464m23 − 6960m3 + 93960) = 0 48h2 + 96h1 + 240h0 + 6m21 − 19m1 − 2m3 + 12 = 0 96h3 + 144h2 + 384h1 + 528h0 + 17m1 − 2m3 − 24 = 0 4h4 + 8h2 + 12h0 − m1 + 2 = 0 4h5 + 8h3 + 12h1 − m1 = 0 h6 − 2h5 + 3h4 − 4h3 + 5h2 − 6h1 + 7h0 = 0 4h7 − 3h6 + 2h5 − h4 + h2 − 2h1 + 3h0 = 0

Length Ten Orthogonal Case: The equations left in the triangular system mentioned in length ten case are: (−3456m81 + 1013760m61 + 36864m51 m3 − 4810752m51 − 331776m41 m3 − 62951424m41 − 5406720m31 m3 + 633249792m31 − 98304m21 m23 + 72990720m21 m3 − 2228699136m21 + 1769472m1 m23 − 257064960m1 m3 + 3460644864m1 − 7864320m23 + 307003392m3 − 1893556224)h1 + 7247757312h30 + (14155776m41 − 2868903936m21 − 75497472m1 m3 + 16590569472m1 + 792723456m3 − 31255953408)h20 + (6912m81 − 3022848m61 − 73728m51 m3 + 17879040m51 + 811008m41 m3 + 297762816m41 + 16121856m31 m3 − 3953049600m31 + 196608m21 m23 − 272990208m21 m3 + 19472203776m21 − 4325376m1 m23 + 1238237184m1 m3 − 45856751616m1 + 9 8 8 23789568m23 − 2047279104m3 + 44454887424)h0 − (108m10 1 − 765m1 − 18m1 m3 − 47448m1 −

1152m71 m3 + 657264m71 + 28992m61 m3 + 2204592m61 + 192m51 m23 + 103488m51 m3 − 106356640m51 + 960m41 m23 − 7099200m41 m3 + 1010070976m41 − 132864m31 m23 + 70831104m31 m3 − 5047712000m31 − 512m21 m33 + 1659648m21 m23 − 331249920m21 m3 + 14768421504m21 + 11264m1 m33 − 7317504m1 m23 + 784624128m1 m3 − 24221738304m1 − 62464m33 + 12056064m23 − 780107904m3 + 17484139008) = 0 1536h2 + 3072h1 + 9216h0 + (3m41 − 836m21 − 16m1 m3 + 6240m1 + 192m3 − 16776) = 0 192h3 + 192h2 + 768h1 + 768h0 + (−39m21 + 494m1 + 4m3 − 2028) = 0 32h4 + 96h2 + 192h0 + (−m21 + 28m1 − 192) = 0 512h5 − 672h4 + 1536h3 − 2016h2 + 3072h1 − 4032h0 + (5m21 − 76m1 ) = 0 912h6 − 2048h5 + 3904h4 − 6144h3 + 8976h2 − 12288h1 + 16128h0 + (−m21 ) = 0 h7 − 3h6 + 6h5 − 10h4 + 15h3 − 21h2 + 28h1 − 36h0 = 0 3h8 − 4h7 + 3h6 − 5h4 + 12h3 − 21h2 + 32h1 − 45h0 = 0 4h9 − 3h8 + 2h7 − h6 + h4 − 2h3 + 3h2 − 4h1 + 5h0 = 0

35

R EFERENCES [1] G. Regensburger and O. Scherzer, “Symbolic computation for moments and filter coefficients of scaling functions,” Annals of Combinatorics, vol. 9, pp. 223–243, 2005. [2] G. Strang and T. Nguyen, Wavelets and Filter Banks. Wellesley, MA: Wellesley-Cambridge Press, 1996. [3] B. Buchberger, “An algorithm for finding the bases elements of the residue class ring modulo a zero dimensional polynomial ideal (german),” PH.D thesis, Univ. of Innsbruck, 1965. [4] N. K. Bose, B. Buchberger, and J. P. Guiver, Multidimentional Systems Theory and Applications. Kluwer Academic Publishers, Dordrecht, The Netherlands, 2003. [5] G. Regensburger, “Parametrizing compactly supported orthonormal wavelets by discrete moments,” to appear in Applicable Algebra in Engineering, Communication and Computing, 2007. [Online]. Available: http://www.ricam.oeaw.ac.at/people/page/regensburger/papers/regensburger06.pdf [6] H. L. Resnikoff, J. Tian, and R. O. Wells, Jr., “Biorthogonal wavelet space: Parametrization and factorization,” SIAM Journal on Mathematical Analysis, vol. 33, no. 1, pp. 194–215, 2002. [7] J. Tian and J. Raymond O. Wells, “A remark of vanishing moments,” In Proc. 30th Asilomar Conf. on Signals, Systems, and Computers, Pacific Grove, CA, pp. 983–987, 1996. [8] A. Fuller, “Root location criteria for quartic equations,” IEEE Transaction on Automatic Control, vol. AC-26, no. 3, pp. 777–782, 1981. [9] I. Daubechies, Ten Lectures on Wavelets. Philadelphia, Pennsylvania: Society for Industrial and Applied Mathematica, 1992. [10] S. Mallat, A Wavelet Tour of Signal Processing. San Diego, CA: Academic Press Inc., 1998. [11] M. Vetterli and J. Kovacevic, Wavelet and Subband Coding. Upper Saddle River, NJ: Prentice Hall PTR, 1995. [12] W. Sweldens and R. Piessens, “Quadrature formulae and asymptotic error expansions for wavelet approximations of smooth functions,” SIAM Journal on Numerical Analysis, vol. 31, no. 4, pp. 1240–1264, 1994. [13] M. Vetterli and C. Herley, “Wavelets and filter banks: Theory and design,” IEEE Trans. Signal Proc., vol. 40(9), pp. 2207–2232, September 1992. [14] M. Vetterli, “Filter banks allowing perfect reconstruction.” Signal Proc., vol. 10(3), pp. 219–244, April 1986. [15] Y. Zhao and M. Swamy, “Design of least asymmetric compactly supported orthogonal wavelet via optimization,” Proc. of the IEEE CCECE, pp. 817–820, 1999. [16] N. Ahuja, S. Lertrattanapanich, and N. K. Bose, “Properties determining choice of mother wavelet,” IEE Proceedings on Vision, Image and Signal Processing, vol. 152, pp. 659–664, 2005.