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Parametrized BiorthogonalWavelets and FIR Filter Bank
Design with Gröbner Bases
M. Peng, N.K. Bose
RICAM-Report 2007-18
1
Parametrized Biorthogonal Wavelets and FIR FilterBank Design with Grobner Bases
MinHsuan Peng and N. K. Bose
Abstract
This paper builds upon the recent results of Regensburger and Scherzer on parametrization of orthonormal wavelets bydiscrete moments (and, therefore, also continuous moments) of scaling function followed by the solution of a parametrized setof polynomial equations in the FIR filter coefficients using Grobner bases. First, parametrization of orthonormal filters with twodiscrete moments as parameters, instead of just one is considered followed by the generalization of the results to the case ofbiorthogonal wavelets. A characterization theorem for the continuous moments of the scaling function and its dual is presentedand proved for biorthogonal wavelets and the result of Regensburger and Scherzer in the orthogonal case is shown to emerge as aspecial case of this. Like in Regensburger’s subsequent work, complete solution to parametrized filters for both the two-parameterorthogonal case and the biorthogonal case (especially for the practically important linear phase biorthogonal class) for filters ofvarying 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, Grobner 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 byexpressing thenth continuous moment as a polynomial (related to a class of distinguished polynomials, called Bell polynomials)of the firstn discrete moments and vice versa. Subsequently, they parametrized the filter coefficients in the dilation equationsgenerating the scaling equation, and compactly supported orthonormal wavelets with several vanishing moments, in termsof the discrete (and, consequently, the continuous) moments. By giving up certain vanishing moments to obtain additionaldegrees of freedom, they were able to generate a parametrized solution set in contrast to previous methods on symboliccomputation of wavelets coefficients which calculated a finite number of solutions that also used algorithmic algebraic toolslike Grobner bases [3] [4, Chapter 4]. Subsequently, Regensburger [5] presented new simplified parametrizations and discussedcomputational aspects for up to ten filter coefficients and at least four vanishing moments with a single discrete momentas parameter. These results in [1] and [5] are significant because previous approaches to parametrization expressed the filtercoefficients in terms of trigonometric functions (and not polynomials) which required the solving of transcendental constraintsfor 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 hasbeen presented recently [6]. Construction of symmetric biorthogonal wavelets and smoothness estimates of such wavelets stillremain largely unexplored. Such problems are addressed here through parametrization by discrete moments. It is known thatvanishing moments of scaling and wavelet functions have different roles. Vanishing moments of the wavelet function is relatedto the wavelet orthogonal projection, while vanishing moments of the scaling function is connected to the wavelet samplingapproximation inside the projection subspace [7]. Here, attention is given to both and solutions for the parametrized FIR (finiteimpulse response) polynomial filter are obtained by the symbolic computer algebra software SINGULAR used to constructGrobner 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 inSection 4 and concluding remarks are summarized in Section 5.
II. T WO PARAMETERS PARAMETRIZATION OF COMPACTLY SUPPORTEDORTHOGONAL WAVELETS
The maximum numberp of vanishing momentsVl, l = 0, 1, ..., p− 1 of the waveletψ(x),
Vl =∫
xlψ(x) = 0
Department of Electrical Engineering, The Pennsylvania State University, University Park, PA 16802, USA E-Mail: [email protected] 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, JohannesKepler University, Linz, Austria) under the scientific direction of Professor Bruno Buchberger.
2
leads to a maximally flat FIR filter of length2p that can be used to generate the Daubechies waveletdbp. Therefore, a lengthten filter generates the compactly supported orthonormal Daubechies waveletdb5, i. e. p = 5.
By giving up the highest vanishing momentVp−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 momentsmk, k = 0, 2, 4, ... of the associated scaling function can be determined by the odd-indexed discrete momentsmk, k = 1, 3, 5, ...up to the number of vanishing moments of the wavelet. By this result, the discrete momentsm1,m2,m3,m4 for a length 10filter, after giving up 1 vanishing moment are related as follows [1]:
m2 =12m2
1
m4 = −38m4
1 + 2m1m3,
Using discrete momentsm1 andm3 as parameters to parameterize a length ten filter with Grobner bases, the first polynomialobtained is only in terms ofm1 andm3. If only one degree of freedom is chosen for parametrization as done in [5], thenm1
andm3 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 momentV0 is a requirementfor the resolution of identity in the theory of wavelets and cannot be eliminated. Therefore, for parametrization with twoparameters (obtained by giving up the highest two vanishing moments) the minimum filter length is six. Therefore, two degreesof freedom for parametrization will be considered by giving up the highest two vanishing moments starting from FIR filtersof length six up to length twelve in increments of two.
A. Setup:
The linear system used to parameterize the filter coefficients contains the equation of normalization (obtained from thescaling functionφ(x) =
∑hkφ(x− k)),
N∑
k=0
hk = 2, (1)
the equation from discrete moments of scaling function,
mn =N∑
k=0
hkkn, (2)
and the sum rule equation, obtained from vanishing moments of wavelet functionψ(x) =∑
(−1)khN−kψ(x− k),2n−1∑
k=0
(−1)n−k(n− k)lhk = 0. (3)
By orthonormality property of scaling function,∫
φ(x)φ(x− l) = δ0,l, (4)
one can obtain the quadratic equations:
N∑
k=1−N
hkhk−2l = 2δ0,l, l = 0, ..., N − 1. (5)
B. Length Six Case
For parametrization with two parameters (for brevity, two-parametrization), give up one more vanishing moment in compar-ison to the one-parametrization case, which means that the only vanishing moment is the zeroth order moment
∫ψ(x) = 0,
and it is equivalent to (from sum rule defined in Eq.(3))
N∑
k=0
(−1)khk = 0. (6)
3
The following linear equations result from the normalization equation (1), the first sum rule (6) and two discrete momentsm1
andm2. Note that since there is only one vanishing moment, therefore,m2 is not determined bym1.
h0 + h1 + h2 + h3 + h4 + h5 =2h0 − h1 + h2 − h3 + h4 − h5 =0h1 + 2h2 + 3h3 + 4h4 + 5h5 =m1
h1 + 4h2 + 9h3 + 16h4 + 25h5 =m2 (7)
The quadratic equations are:
h0h2 + h1h3 + h2h4 + h3h5 =0h0h4 + h1h5 =0. (8)
To solve for the six filter coefficients, parametrized bym1 andm2, use SINGULAR to obtain the reduced Grobner basis basedon lexicographical order and rankingh5 > ... > h1 > h0 in the form of a triangular system of polynomial equations:
16384h40 + (28672m1 − 4096m2 − 86016)h3
0 + (22016m21 − 6656m1m2 − 124416m1 + 512m2
2+
18432m2 + 177664)h20 + (8128m3
1 − 3840m21m2 − 67584m2
1 + 608m1m22 + 21248m1m2+
185696m1 − 32m32 − 1696m2
2 − 28832m2 − 170016)h0 + (1300m41 − 864m3
1m2 − 13872m31+
216m21m2
2 + 6904m21m2 + 55224m2
1 − 24m1m32 − 1152m1m2
2 − 18264m1m2 − 97152m1+
m42 + 64m3
2 + 1530m22 + 15904m2 + 64009) = 0 (9)
(192m21 − 80m1m2 − 720m1 + 8m2
2 + 160m2 + 584)h1 − 1024h30 + (−1664m1 + 256m2 + 4608)h2
0+
(−784m21 + 224m1m2 + 4512m1 − 16m2
2 − 640m2 − 6480)h0 + (−50m31 + 14m2
1m2 + 522m21−
m1m22 − 100m1m2 − 1817m1 + 4m2
2 + 176m2 + 2116) = 0 (10)
(96m21 − 40m1m2 − 360m1 + 4m2
2 + 80m2 + 292)h2 + 1024h30 + (1664m1 − 256m2 − 4608)h2
0+
(1168m21 − 384m1m2 − 5952m1 + 32m2
2 + 960m2 + 7648)h0 + (242m31 − 118m2
1m2 − 1986m21+
19m1m22 + 660m1m2 + 5191m1 −m3
2 − 55m22 − 869m2 − 4379) = 0 (11)
4h3 − 8h0 + (−6m1 + m2 + 13) = 0 (12)
(96m21 − 40m1m2 − 360m1 + 4m2
2 + 80m2 + 292)h4 − 1024h30 + (−1664m1 + 256m2 + 4608)h2
0+
(−1072m21 + 344m1m2 + 5592m1 − 28m2
2 − 880m2 − 7356)h0 + (−242m31 + 118m2
1m2 + 1890m21−
19m1m22 − 620m1m2 − 4831m1 + m3
2 + 51m22 + 789m2 + 4087) = 0 (13)
(192m21 − 80m1m2 − 720m1 + 8m2
2 + 160m2 + 584)h5 + 1024h30 + (1664m1 − 256m2 − 4608)h2
0+
(1168m21 − 384m1m2 − 5952m1 + 32m2
2 + 960m2 + 7648)h0 + (338m31 − 182m2
1m2 − 2418m21+
33m1m22 + 860m1m2 + 5753m1 − 2m3
2 − 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 functionsof parametersm1 andm2. Note that the system of equations is a triangular system. The solution to the fourth order polynomialwill depend on the coefficients. So there are regions in the parameter space ofm1, m2 that yield real solutions. Use the rootlocation criterion for the quartic equations (see Fuller[8])
a0x4 + a1x
3 + a2x2 + a3x + a4 = 0 (15)
with coefficients all real anda0 > 0, based on the determinants of a sequence of Sylvester matrices:
∆3 =
∣∣∣∣∣∣
a0 a1 a2
0 4a0 3a1
4a0 3a1 2a2
∣∣∣∣∣∣(16)
∆5 =
∣∣∣∣∣∣∣∣∣∣
a0 a1 a2 a3 a4
0 a0 a1 a2 a3
0 0 4a0 3a1 2a2
0 4a0 3a1 2a2 a3
4a0 3a1 2a2 a3 0
∣∣∣∣∣∣∣∣∣∣
(17)
4
m1
m2
(a)∆3 > 0
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
35
40
m1
m2
(b)∆5 > 0
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
35
40
m1
m2
(c)∆7 > 0
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
35
40
m1
m2
(d)∆3, ∆
5, ∆
7 > 0
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
35
40
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 a1 a2 a3 a4 0 00 a0 a1 a2 a3 a4 00 0 a0 a1 a2 a3 a4
0 0 0 4a0 3a1 2a2 a3
0 0 4a0 3a1 2a2 a3 00 4a0 3a1 2a2 a3 0 0
4a0 3a1 2a2 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 fourthdegree equation will have only two real roots and a pair of complex roots if and only if any one of the following conditionsholds,
∆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 parametersm1 andm2 as axes. In Figure 1(d), the green colored region gives four realsolutions and in Figure 1(c) the white region corresponding to∆7 < 0 gives two complex solutions. The zone with all realroots can be obtained by solving the inequalities∆5 > 0 and∆7 > 0, in this case. One of the boundaries form2 is defined
5
from
m2 >12m2
1, (21)
which is from∆5 and the other, determined by∆7, is defined from
m2 < 5m1 − 4−√−m2
1 + 10m1 − 9. (22)
The range ofm1 extends from5 −√15 to 5 +√
15 and the corresponding range ofm2 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
5m1 − 4−√−m2
1 + 10m1 − 9 < m2 < 5m1 − 4 +√−m2
1 + 10m1 − 9. (23)
The valid intervals form1 is [1, 9] and the correspondingm2 lies in the interval [1, 41]. Inequality (23) is not linear, and themaximum and minimum values ofm2 are41.3961 and0.6039 respectively.
From the above observation, it is interesting to note that the real solution boundary is generated by
m2 =12m2
1,
which is the relation betweenm2 andm1 when there are two vanishing moments. Therefore, when the parameters are closerto this boundary, the corresponding wavelets are smoother.
Special Case
With parametersm1 = 1 and m2 = 1 one obtains the Haar wavelet and a translated one when parameters arem1 = 9m2 = 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 betweenm1 andm2,
m2 =12m2
1,
generates the boundary of real roots region. For the special case of Daubechies wavelet, the third vanishing moment gives onemore linear equation
−9h0 + 4h1 − h2 − h4 + 4h5 = 0. (25)
Substituting the parameterized filter coefficients into equations (24) and (25), one gets two real solution sets form1 andm2
m1 = 1.6348, m2 = 1.3363m1 = 8.3652, m2 = 34.9883.
The first solution generates thedb3 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 determinedby the first discrete moment:
m2 =12m2
1.
The linear equations are
h0 + h1 + h2 + h3 + h4 + h5 + h6 + h7 = 2h0 − h1 + h2 − h3 + h4 − h5 + h6 − h7 = 04h0 − 3h1 + 2h2 − h3 + h5 − 2h6 + 3h7 = 0
h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 = m1
h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 =12m2
1
h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 = m3 (26)
6
m1
m2
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
35
40
Fig. 2. Special case of filter coefficients length six. The bold line is defined by the relationm2 = 1/2m21 and the black point in the left corner is the first
parameter set (1.6348, 1.3363) that givesdb3 and the translated version is the other point (8.3652, 34.9883) in the right corner.
and the quadratic equations (from orthogonality) are:
h0h2 + h1h3 + h2h4 + h3h5 + h4h6 + h5h7 = 0h0h4 + h1h5 + h2h6 + h3h7 = 0h0h6 + h1h7 = 0 (27)
To parametrize eight filter coefficients, apply SINGULAR with lexicographical ordering and rankingh7 > ... > h1 > h0 to thesets of equations defined in (26) and (27) and obtain the reduced Grobner basis as a triangular system of polynomial equations.The first polynomial withh0 as indeterminate and coefficients parametrized bym1 andm3 is:
1358954496h40 + (212336640m2
1 − 707788800m1 − 56623104m3 − 84934656)h30 + (20570112m4
1−177389568m3
1 − 9732096m21m3 + 518750208m2
1 + 40108032m1m3 − 641433600m1 + 1179648m23−
19464192m3 + 368934912)h20 + (922752m6
1 − 12524544m51 − 663552m4
1m3 + 67977216m41+
6100992m31m3 − 187077120m3
1 + 156672m21m2
3 − 20173824m21m3 + 266489856m2
1 − 718848m1m23+
29601792m1m3 − 167380992m1 − 12288m33 + 681984m2
3 − 18247680m3 + 23639040)h0 + (26325m81−
538488m71 − 23328m6
1m3 + 4830624m61 + 357264m5
1m3 − 24799104m51 + 7776m4
1m23 − 2284416m4
1m3+
79502620m41 − 79488m3
1m23 + 7778048m3
1m3 − 162536640m31 − 1152m2
1m33 + 307104m2
1m23−
14686848m21m3 + 206735040m2
1 + 5888m1m33 − 533376m1m2
3 + 14114880m1m3 − 150128640m1+
64m43 − 7680m3
3 + 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 lengthythey are omitted here but can be found in Appendix. After solving (28) forh0, substitute this value into the next equation andsolve. Continue the process to generate the complete solution set of coefficients. Similar to the six coefficients case, there arealso certain regions defined by the valid ranges ofm1 and m3. Using the root location criterion (19) and (20), get all realsolutions 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.
7
m1
m3
(a) ∆3 > 0
1 2 3 4 5 6 7−40
−30
−20
−10
0
10
20
30
40
m1
m3
(b) ∆5 > 0
1 2 3 4 5 6 7−40
−30
−20
−10
0
10
20
30
40
m1
m3
(c) ∆7 > 0
1 2 3 4 5 6 7−40
−30
−20
−10
0
10
20
30
40
m1
m3
(d)∆3, ∆
5, ∆
7 > 0
1 2 3 4 5 6 7−40
−30
−20
−10
0
10
20
30
40
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 arenot 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 rankingh7 > ... > h1 > h0 > m3 to the preceding equation andthe sets of equations defined in (26) and (27), the first polynomial, in the Grobner base, is a polynomial in the variablem3,with its coefficients parametrized bym1, as given next.
(−256m21 + 3584m1 − 12800)m2
3 + (96m51 − 672m4
1 − 7936m31 + 81536m2
1−190208m1 + 102144)m3 + (−9m8
1 + 1488m61 − 4368m5
1 − 66400m41 + 443072m3
1−1059584m2
1 + 1180032m1 − 544320) = 0, (29)
and this is plotted in Figure 4. When the pointm1,m3 is close to the real-valued solution set for the preceding equation, theassociated 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 momentV3 yields
−27h0 + 8h1 − h2 + h4 − 8h5 −+27h6 − 64h7 = 0.
The polynomial equation inm3 obtained by applying SINGULAR with lexicographical ordering and rankingh7 > ... > h1 >h0 > m1 > m3 on the preceding two equations and the sets of equations defined in (26) and (27) is
m83 − 980m7
3 + 362376m63 − 68283068m5
3 + 6824666674m43 − 362290145364m3
3+
11165951983816m23 − 191266139512700m3 + 94299960728785 = 0.
The solution that yieldsdb4 corresponds to
m3 = 0.50784, m1 = 2.0108.
8
m1
m3
1 2 3 4 5 6 7−40
−30
−20
−10
0
10
20
30
40
Fig. 4. Special case of filter coefficients of length eight. The bold line is defined by the relation betweenm3 andm1 described by Eq.(29) and the blackdot is the first parameter set (2.0108, 0.50784) that givesdb4.
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 thehighest two vanishing moments. The linear equations are
h0 + h1 + h2 + h3 + h4 + h5 + h6 + h7 + h8 + h9 = 2h0 − h1 + h2 − h3 + h4 − h5 + h6 − h7 + h8 − h9 = 0
−5h0 + 4h1 − 3h2 + 2h3 − h4 + h6 − 2h7 + 3h8 − 4h9 = 0−25h0 + 16h1 − 9h2 + 4h3 − h4 − h6 + 4h7 − 9h8 + 16h9 = 0
h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 + 8h8 + 9h9 = m1
h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 + 64h8 + 81h9 =12m2
1
h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 + 512h8 + 729h9 = m3, (30)
and the quadratic equations (from orthogonality) are
h0h2 + h1h3 + h2h4 + h3h5 + h4h6 + h5h7 + h6h8 + h7h9 = 0h0h4 + h1h5 + h2h6 + h3h7 + h4h8 + h5h9 = 0h0h6 + h1h7 + h2h8 + h3h9 = 0h0h8 + h1h9 = 0. (31)
As in the length eight case, use reduced Grobner bases with lexicographic order and rankingh9 > ...h1 > h0 usingm1 andm3 as parameters. In the resulting triangular system of equations, the first equation only involvesh0 as indeterminate withcoefficients parametrized bym1 andm3. After solving the first equation forh0, substituteh0 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 herebut 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
9
m1
m3
(a) ∆3 > 0
0 1 2 3 4 5 6 7 8 9 10−50
0
50
100
150
200
250
300
m1
m3
(b) ∆5 > 0
0 1 2 3 4 5 6 7 8 9 10−50
0
50
100
150
200
250
300
m1
m3
(c) ∆7 > 0
0 1 2 3 4 5 6 7 8 9 10−50
0
50
100
150
200
250
300
m1
m3
(d) ∆3, ∆
5, ∆
7 > 0
2 3 4 5 6 7 8 9
0
50
100
150
200
250
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 theothers, 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 ofcomputing. 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 ofm1andm3 after applying SINGULAR with lexicographic order and rankingh9 > ...h1 > h0 > m3 on the system of equationsabove:
1024m43 + (−3072m3
1 + 55296m21 − 489472m1 + 1419264)m3
3 + (2880m61 − 93312m5
1+
1540224m41 − 15303168m3
1 + 97677312m21 − 358511616m1 + 548785152)m2
3 + (−864m91+
31104m81 − 496512m7
1 + 3768768m61 − 4056192m5
1 − 176214528m41 + 1512364544m3
1−5357366784m2
1 + 8252955648m1 − 4229148672)m3 + (81m121 − 2916m11
1 + 40716m101 −
155520m91 − 2354328m8
1 + 31658688m71 − 102669504m6
1 − 590398848m51 + 6210049216m4
1−22429995264m3
1 + 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 relationbetweenm1 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 waveletdb5 has maximum number of vanishing moments: five. The linear
10
m1
m3
2 3 4 5 6 7 8 9
0
20
40
60
80
100
120
140
160
180
200
Fig. 6. Special case of filter coefficients length ten. The bold line defines relation betweenm3 andm1 described in equation (29) and the black point is theparameter set (2.3878, 1.7018) that givesdb5.
equation given by the fifth vanishing momentV4 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 form3 and an equationrelatingm1 in terms ofm3. These two equations give four real and twelve complex roots. The first real root
m3 = 1.7018,m1 = 2.3878
gives Daubechies waveletdb5 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], thefourth discrete momentm4 can be determined by the odd-indexed momentsm1 andm3, in this case,
m4 = −38m4
1+2m1m3.
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 = 0
−25h0 + 16h1 − 9h2 + 4h3 − h4 − h6 + 4h7 − 9h8 + 16h9 − 25h10 + 36h11 = 0
−125h0 + 64h1 − 27h2 + 8h3 − h4 + h6 − 8h7 + 27h8 − 64h9 + 125h10 − 216h11 = 0
h1 + 2h2 + 3h3 + 4h4 + 5h5 + 6h6 + 7h7 + 8h8 + 9h9 + 10h10 + 11h11 = m1
h1 + 4h2 + 9h3 + 16h4 + 25h5 + 36h6 + 49h7 + 64h8 + 81h9 + 100h10 + 121h11 =1
2m2
1
h1 + 8h2 + 27h3 + 64h4 + 125h5 + 216h6 + 343h7 + 512h8 + 729h9 + 1000h10 + 1331h11 = m3
h1 + 16h2 + 81h3 + 256h4 + 625h5 + 1296h6 + 2401h7 + 4096h8 + 6561h9 + 10000h10+14641h11
= −3
8m4
1+2m1m3
(34)
and the quadratic equations (from orthogonality) are
h0h2 + h1h3 + h2h4 + h3h5 + h4h6 + h5h7 + h6h8 + h7h9 + h8h10 + h9h11 = 0
h0h4 + h1h5 + h2h6 + h3h7 + h4h8 + h5h9 + h6h10 + h7h11 = 0
h0h6 + h1h7 + h2h8 + h3h9 + h4h10 + h5h11 = 0
h0h8 + h1h9 + h2h10 + h3h11 = 0
h0h10 + h1h11 = 0. (35)
Applying SINGULAR with lexicographical ordering and rankingh11 > ... > h1 > h0 to the above sets of equations, the firstpolynomial equation in the triangular system of polynomial equations in the reduced Grobner basis is an equation of fourthdegree inh0. As in the previous cases, the root location criteria in (19) and (20) can be used. The complex set of equationsare omitted here. The zones∆3 > 0, ∆5 > 0 and∆7 > 0 are presented in Figures 7(a),(b),(c), respectively, and (d) definesthe region∆3, ∆5, ∆7 > 0.
Special CaseIf the fifth vanishing momentV4 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 rankingh11 > ... > h1 > h0 > m3 to the preceding equation andthe sets of equations defined in (34) and (35), the first polynomial, in the reduced Grobner basis, is a polynomial in variablem3, given by
c0m63 + c1m
53 + c2m
43 + c3m
33 + c4m
23 + c5m3 + c6 = 0, (37)
where the coefficientcsk are parametrized bym1. 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 momentV5 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 fordb6. Applying SINGULAR with lexicographicalordering and rankingh11 > ... > h1 > h0 > m1 > m3, the first polynomial in the Grobner basis is of 32nd degree inm3.Solve this equation form3 and substitute into the preceding relation (37). Then solve form1. 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 BIORTHOGONAL WAVELETS
In the previous section, the parametrization is on FIR filters of the compactly supported orthonormal wavelets. The nonlin-earity of phase response of those filters is undesirable in many applications. However, FIR filters that generate orthonormalwavelets cannot have linear phase except in the case of Haar wavelets (see [9, page.253-255] ). Therefore, parametrization oflinear phase FIR filters capable of generating biorthogonal wavelets is considered. For a start, either one or two degrees offreedom 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 detailson biorthogonal wavelets, refer to [9],[10] and [11].
12
m1
m3
(a) ∆3 > 0
1 2 3 4 5 6 7
0
20
40
60
80
100
120
m1
m3
(b) ∆5 > 0
1 2 3 4 5 6 7
0
20
40
60
80
100
120
m1
m3
(c) ∆7 > 0
1 2 3 4 5 6 7
0
20
40
60
80
100
120
m1
m3
(d) ∆3, ∆
5, ∆
7 > 0
3 3.5 4 4.5 5 5.5 6 6.5 7 7.50
20
40
60
80
100
120
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, andcase (d) includes the region∆3,∆5,∆7 >0 (only the subregion pointed by arrow in the middle.) The numerical computation becomes much more sensitiveto the variation of parameters in comparison to lower length cases.
A. A review of Biorthogonal Wavelets
For biorthogonal wavelets, the dual scaling functionφjk = 2j/2φ(2j − k) and dual mother waveletψjk = 2j/2ψ(2j − k)which constitute the dual subspacesVj andWj are, respectively, orthogonal to the waveletψjl and the scaling functionφjl,i.e.
〈φjk, ψjl〉 = 0, 〈ψjk, φjl〉 = 0. (38)
The scaling functionφjk is orthogonal to its dualφjl, i.e
〈φjk, φjl〉 = δkl, (39)
and so is the wavelet functionψjl to its dualψj′l, i.e
〈ψjk, ψj′l〉 = δjj′δkl. (40)
The properties (38), (39) and (40) are also known as biorthogonality. A scaling function and a mother wavelet satisfy thedilation equation and the wavelet equation, respectively, i.e.
φ(x) =N∑
k=0
hkφ(2x− k) and ψ(x) =N∑
k=0
gkφ(2x− k) (41)
φ(x) =N∑
k=0
hkφ(2x− k) and ψ(x) =N∑
k=0
gkφ(2x− k) (42)
13
m1
m3
1 2 3 4 5 6 7
0
20
40
60
80
100
120
Fig. 8. Special case of filter coefficients length twelve. The bold line is the relation betweenm3 andm1 described by Eq. (29), and the black dot is theparameter set (2.7643, 3.382) that givesdb6.
whereN can beodd or even. The relationships betweengk and hk, and gk andhk, obtained through the condition of exactreconstruction (refer to [9, p.262-263] (8.3.1) and (8.3.2)), are given as
gk = (−1)k+1hN−k (43)
gk = (−1)k+1hN−k, (44)
whereN = 2n− 1 in the [9]. Therefore, wavelet functions can be written as
ψ(x) =N∑
k=0
(−1)k+1hN−kφ(2x− k) (45)
and
ψ(x) =N∑
k=0
(−1)k+1hN−kφ(2x− k). (46)
Substituteφ(x) and φ(x) defined in (41) and (42) into orthogonality relation betweenφ andφ, i.e,∫
φ(x)φ(x− l)dx = δ0,l (47)
to obtainN∑
k=0
hkhk = 2 (48)
and a homogeneous equation
N∑
k=0
hkhk−2l = 0. l = 1, ..., (N − 1)/2. (49)
14
The filter coefficientshi and its dualhi satisfy the normalization constraints:(refer to Eq.(1))
N∑
k=0
hk = 2 andN∑
k=0
hk = 2. (50)
Next, reduced vanishing momentsfor the case of symmetric filter coefficients are introduced. For convenience and clarity, theeven 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∫
xlψ(x)dx = 0, l = 0, 1, ..., p− 1. (51)
Substitute the wavelet function in (45) into this condition to obtain the sum rule
N∑
k=0
(−1)kklhk = 0, l = 0, ..., p− 1, (52)
which is equivalent to (see [1])
2n−1∑
k=0
(−1)n−k(n− k)lhk = 0, N = 2n− 1. (53)
Since the filter coefficientsgk of ψ are related tohk, the vanishing moments ofψ generate the sum rule ofhk. Furthermore,since the filter coefficients are symmetric (linear phase), that is,
hk = hN−k, k = 0, 1, ..., (N − 1)/2, (54)
equation (53) can be rewritten as
n−1∑
k=0
(−1)k[(n− k)l − (k + 1− n)l]hk = 0. (55)
Note that the left hand side of Eq.(55) is identically zero whenl = 0, implying that the first sum rule holds trivially. It can beproved by mathematical induction that (55), for odd values of integerl, becomes
n−1∑
k=0
(−1)k[(n− k)l+1 − (k + 1− n)l+1]hk = 0, l = 1, 3, 5, 7... (56)
From Eqs.(55) and (56), for a linear phase FIR filter of even length, thelth sum rule is the same as the(l + 1)th sum ruleprovidedl = 1, 3, 5... is odd.
Odd length (N even)The length of FIR filters associated with biorthogonal wavelets do not have to beevenbecause doubleshift orthogonality does not hold anymore, i.e,
N∑
k=0
hkhk−2l 6= 0.
If N is even(length isodd), the sum rule obtained in equation (53) has to be modified as
2n∑
k=0
(−1)n−k+1(n− k + 1)lhk = 0, N = 2n. (57)
Using symmetry property
hk = hN−k, k = 0, 1, ..., N/2, (58)
the sum rule (57) becomes
n−1∑
k=0
(−1)k+1[(n− k + 1)l + (k + 1− n)l]hk − hn = 0, (59)
15
By mathematical induction one can prove, whenl = 0, 2, 4, 6..., that Eq. (59) reduces to
n−1∑
k=0
(−1)k+1[(n− k + 1)l+1 + (k + 1− n)l+1]hk − hn = 0. (60)
From Eqs.(59) and (60), for a linear phase FIR filter of odd length, thelth sum rule is the same as the(l + 1)th sum ruleprovidedl = 0, 2, 4... is even.
The preceding discussions lead to the notion ofreduced vanishing momentsfor symmetric biorthogonal wavelets. Since thesymmetry property of filter coefficients makes thelth vanishing moment produce the same sum rule as the(l + 1)th, definethe numberp of reduced vanishing momentsas follows.
Definition 1: In the linear phase biorthogonal wavelet system
ψ(x) =N∑
k=0
gkψ(2x− k),
the wavelet functionψ hasp reduced vanishing momentsVl if it satisfies
Vl =∫ ∞
−∞xlψ(x)dx = 0,
{for l = 0, 1, ..., 2(p− 1) when N is odd;for l = 0, 1, ..., 2p− 1 when N is even.
B. Discrete and Continuous MomentsDiscrete Moments for Even Length
The discrete moment of the analysis filter bank scaling function is defined asmn =∑N
k=0 hkkn, N is odd. By symmetry,hk = hN−k, implying that the discrete moment beomes
mn =
N−12∑
k=0
(kn + (N − k)n)hk. (61)
In particular,
m0 =N∑
k=0
hk = 2, i.e
N−12∑
k=0
hk = 1 (62)
m1 =N∑
k=0
hkk = N
N−12∑
k=0
hk = N (63)
m2 =
N−12∑
k=0
(k2 + (N − k)2)hk. (64)
Define qn = kn + (N − k)n. Thenq0 = 2, q1 = N and q2 = k2 + (N − k)2 etc.. Solve fork in terms ofq2 and substituteinto qn, n = 3, 4, ..., 7 to get,
q3 =12(3Nq2 −N3) (65)
q4 =12(q2
2 + 2N2q2 −N4) (66)
q5 =14(5Nq2
2 −N5) (67)
q6 =14(q3
2 + 6N2q22 − 3N4q2) (68)
q7 =18(7Nq3
2 + 7N3q22 − 7N7q2 + N7) (69)
From (66) and (68) one can write
q22 =2q4 − 2N2q2 + N4, (70)
q23 =4q6 − 6N2q2
2 + 3N4q2. (71)
16
Substituteq22 andq3
2 into (67) and (69) to representq3, q5 andq7 as
q3 =12(3Nq2 −N3)
q5 =14(10Nq4 − 10N3q2 + 4N5)
q7 =18(28Nq6 − 70N3q4 + 84N5q2 − 34N7)
From (62) and (63),
m0 = 2, m1 = N.
Substituteq3, q5 andq7 into (61) to get the odd-indexed moments in terms of the smaller even-indexed moments, noting thatm1 = N is a constant.
m3 =12(3m1m2 −m3
1) (72)
m5 =14(10m1m4 − 10m3
1m2 + 4m51) (73)
m7 =18(28m1m6 − 70m3
1m4 + 84m51m2 − 34m7
1). (74)
Discrete Moments for Odd Length
For the odd length filter coefficient case, Eqs. (72), (73) and (74) still hold. With symmetryhk = hN−k, in the N even case,here, the discrete moments are rewritten as
mn =N∑
k=0
hkkn =
N2 −1∑
k=0
(kn + (N − k)n)hk + (N
2)nhN/2, N is even.
As in the even length case, letqn = kn + (N − k)n and obtain the sameq3, q4, ..., q7 as described above. Substituteq3 =12 (3Nq2 −N3) into
m3 =
N2 −1∑
k=0
(k3 + (N − k)3)hk + (N
2)3hN/2,
to get
m3 =3m1m2 −m3
1
2The expressions form5,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 lengthcases. Next the relationship between continuous and discrete moments of the scaling function and its dual is investigated.
The Relationship Between Discrete and Continuous Moments for Biorthogonal wavelet
The continuous and discrete moments of the dual scaling functionφ(x) are defined, respectively, by
Mn =∫
xnφ(x)dx (75)
mn =∑
k
hkkn. (76)
They also satisfy the recursive equation obtained by Strang & Nguyen[2, p.396] and [1]:
Mn =1
2n+1 − 2
n∑
i=1
(ni
)miMn−i (77)
mn = (2n+1 − 2)Mn −n−1∑
i=1
(ni
)miMn−i. (78)
From the proved results in [1], ifp moments of the waveletψ vanish, then∑
k
(x− k)nφ(x− k) = Mn, for 0 ≤ n ≤ p− 1. (79)
17
and∑
k
knφ(x− k) =n∑
i=0
(ni
)(−1)ixn−iMi, for 0 ≤ n ≤ p− 1. (80)
If there arep vanishing moments for the dual waveletψ, thenφ andMn in the above equations (79) and (80) will be replacedby φ andMn. For the proof, see Sweldens &Piessens[12]. With the above knowledge a relationship between dual continuousmoments of biorthogonal scaling functions can be obtained as the following.
Theorem 1:Let φ, φ ∈ L2(R) be a scaling functions and its dual with biorthogonality< φ(x)φ(x−k) >k∈Z= 0. Let p ∈ Nbe odd, and the associated wavelet functionψ hasp + 1 vanishing moments. Then
Mp+1 + Mp+1 =p∑
i=1
(−1)i+1
(pi
)(MiMp−i+1 + MiMp−i+1). (81)
Proof:Inspired by [1], define
sk ,< x, φ(x)φ(x− k) > for k ∈ Z. (82)
By biorthogonality,s−k can be written as
s−k =< x, φ(x)φ(x + k) >=< x− k, φ(x− k)φ(x) >=< x, φ(x− k)φ(x) >= sk. (83)
Furthermore, sincep is odd∞∑
k=−∞kpsk +
∞∑
k=−∞kps−k
=∞∑
k=0
kpsk +0∑
k=−∞kpsk +
∞∑
k=0
kps−k +0∑
k=−∞kps−k
=∞∑
k=0
kpsk −∞∑
k=0
kps−k +∞∑
k=0
kps−k −∞∑
k=0
kpsk
= 0. (84)
Using (82), (80) and the definition of continuous moment, write∞∑
k=−∞kpsk =< x, φ(x)
∞∑
k=−∞kpφ(x− k) >=
p∑
i=0
(pi
)(−1)iMi < x, φ(x)xp−i >=
p∑
i=0
(pi
)(−1)iMiMp−i+1. (85)
Furthermore, sinces−k can be written as the dual ofsk, derived in (83). Therefore,∞∑
k=−∞kps−k =
∞∑
k=−∞kpsk =
p∑
i=0
(pi
)(−1)iMiMp−i+1. (86)
Finally, from (84) along with the above results (85) and (86),∞∑
k=−∞kpsk +
∞∑
k=−∞kps−k =
p∑
i=0
(pi
)(−1)i(MiMp−i+1 + MiMp−i+1) = 0. (87)
For example, setM0 = M0 = 1, and whenp = 1, 3, 5, there arep + 1 = 2, 4, 6 vanishing moments in the wavelet functionψ.
M2 + M2 =2M1M1 (88)
M4 + M4 =4(M1M3 + M1M3)− 6M2M2 (89)
M6 + M6 =6(M1M5 + M1M5)− 15(M2M4 + M2M4) + 20M3M3 (90)
Special Case: Linear PhaseSubstitute 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 precedingexamples to get
m2 + m2 =m21 (91)
m4 + m4 =m41 − 3m2
1m2 + 3m22 (92)
2m6 + 2m6 =2m61 − 15m4
1m2 + 45m21m
22 − 45m3
2 − 15m21m4 + 30m2m4 (93)
Orthogonal CaseIf φ = φ, then Eq.(83) will becomesk = s−k andMn = Mn. Equations (88) to (90) becomes
M2 =M21 (94)
M4 =− 3M22 + 4M1M3 (95)
M6 =10M23 + 6M1M5 − 15M2M4. (96)
SubstituteM2 in the above Eq.(94) into the other two equations to obtain
M4 =− 3M41 + 4M1M3 (97)
M6 =10M23 + 6M1M5 − 60M3
1 M3 + 45M61 , (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 caseswill be discussed. Furthermore, the filter of dual scaling functionφ does not have to be the same length withφ. Vetterli &Herley[13] showed that in two channel, perfect reconstruction filter bank, where all filters have linear phase, the two analysisfilters (φ andψ) 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 functionψ and ψ can be determined once one has scaling functionφ andφ, since the filter coefficientsgk of wavelet functionψ can be determined byhk and coefficientsgk of dual wavelet functionψ can be determined byhk:
gk = (−1)k+1hN−k and gk = (−1)k+1hN−k, (99)
where support k is[0, N ]. The length condition ofhk is thus the same withgk which is one of the above cases.The parametrization of wavelets will become the parametrization of scaling functionφ and its dual functionφ. In the
following, parametrization by different length types ofφ and φ (noted asφ/φ) will be discussed. One will be even/even andthe 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 theallocation of reduced vanishing moments. For a maximally flat FIR filter that is associated with a symmetric biorthogonalwavelet, the number of reduced vanishing moments isp = L/2, when L = N + 1 is the filter length for oddN . For oddlength case,p = (L− 1)/2.
The first reduced vanishing momentV0 is a requirement for theresolution of identityin the theory of wavelets and cannot beeliminated. However, the symmetric coefficients will make the first sum rule redundant for length even case. Therefore, in thefollowing length even case parametrization, the first sum rule is omitted. (Note that for length odd case, the first sum rule isstill useful.) The minimum number of reduced vanishing moments for bothφ and φ is 1 for even as well as odd length cases.Therefore, the number of degrees of freedom available for allocation isp−1 for length even case andp−2 for length odd case.For a length six symmetric biorthogonal wavelet, the maximum number of reduced vanishing moments is6/2 = 3. The numberof degrees of freedom to allocate them isp− 1 = 2 implying that (1,3) and (2,2) are the two possible distributions of reducedvanishing moments in(φ, φ). For a length five case, the maximum number of reduced vanishing moments is(5− 1)/2 = 2.Here the number of degrees of freedom to allocate them isp− 2 = 0 which means there can only be (1,1) reduced vanishingmoments in(φ, φ).
Below, the highest two reduced vanishing moments,Vp−1 and Vp−2, are given up to obtain either one or two degrees offreedom of parameters. For example, for a length eight case parametrization with one parameter,p = 3. The number of degreesof freedom in their allocation isp− 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 systemsatisfy the normalization constraint (50):
N∑
k=0
hk = 2 andN∑
k=0
hk = 2.
The orthogonality of integer translates of scaling functionφ(x) andφ(x − l) does not hold in the biorthogonal case, andtherefore the length of filter coefficients is not constrained to be even. The quadratic equation which comes from orthogonalityis also not satisfied anymore. The biorthogonality betweenφ(x) and φ(x) reflected in Eqs.(48) and (49) are rewritten below,for convenience.
N∑
k=0
hkhk = 2 (100)
andN∑
k=0
hkhk−2l = 0. while l = 1, ..., (N − 1)/2. (101)
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 forφ becomes
N−12∑
k=0
hk = 1, (102)
and for φ the normalization equation becomesN−1
2∑
k=0
hk = 1. (103)
Furthermore, the sum rule leads ton−1∑
k=0
(−1)n−k[(n− k)l − (k + 1− n)l]hk = 0, (104)
whereN = 2n− 1 and l = 1, ..., p− 1, p is the number of vanishing moments ofψ.
4/4 Case
When the length of analysis filter coefficients of symmetric biorthogonal wavelet is four , one obtains one parameter by givingup the second reduced vanishing moment,V1. The remaining reduced vanishing moment isV0. The sum rule is eliminated.Therefore, the linear system obtained from Eqs.(102),(104) and discrete moment are
h0 + h1 = 19h0 + 5h1 = m2.
Note that from Eqs.(62) and (63)m0 = 2, m1 = N . Solve this linear system to get
h0 =m2 − 5
4
h1 =−m2 + 9
4. (105)
20
The linear and quadratic equations associated with dual filterh are from Eqs. (103) and (101):
h0 + h1 = 1
h0h0 + h1h1 = 1
h0h1 + h1h0 = 0. (106)
Substitute the result of (105) into above dual filter equations to obtain
h0 =m2 − 5
2(m2 − 7)
h1 =m2 − 9
2(m2 − 7). (107)
One more reduced vanishing moment from Eq. (51) withl = 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
h0 = −0.5, h1 = 1.5, h2 = 1.5, h3 = −0.5,
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,V2, onedegree of freedom for parametrization is obtained and the remaining numberp = 2. Since the minimum number of reducedvanishing moments for each of the scaling function filters is one, the only way to distribute the reduced vanishing momentsis 1 in φ and 2 inφ .
Furthermore, if the highest two reduced vanishing moments are given up to obtain two degrees of freedom, then there isonly one reduced vanishing moments left for the analysis filter or the synthesis filter. In the following, these conditions arediscussed separately.
Reduced Vanishing moments(1,2)
The two reduced vanishing moments in the dual wavelet functionψ gives one sum rule forhk. The linear system forparametrization of analysis filter coefficients becomes
h0 + h1 + h2 = 1−5h0 + 3h1 − h2 = 0
25h0 + 17h1 + 13h2 = m2.
This system is solved for theh′ks :
h0 =m2 − 14
16
h1 =m2 − 10
16
h2 =−m2 + 20
8. (108)
For the dual filter coefficientsh, the linear and quadrature equations are
h0 + h1 + h2 = 1
h0h0 + h1h1 + h2h2 = 1
h2h0 + h2h1 + (h0 + h1)h2 = 0. (109)
21
Substitute (108) into the above system to solve forhk :
h0 =−m2
2 + 26m2 − 1688(m2 − 16)
h1 =m2
2 − 22m2 + 1208(m2 − 16)
h2 =m2 − 20
2(m2 − 16).
As a special case, add one reduced vanishing moment in dual wavelet functionψ to obtain a linear independent sum rule withl = 3,
−35h0 + 9h1 − h2 = 0.
Substitute equation (108) into above sum rule to getm2 = 15 and
hk = [116
,516
,58,
58,
516
,116
]
hk = [38, −15
8,
52,
52, −15
8,
38],
which corresponds to the maximum number of reduced vanishing moments case. Since the numerator ofh0 and h1 in theparametrization have a common rootm2 = 12, therefore
hk = [−18,
18, 1, 1,
18, −1
8]
hk = [0, 0, 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 evenmultiple 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 discretemomentm3 is dependent onm2 (see 72), the fourth discrete momentm4 in the analysis filter orm2 in the synthesis filtercan 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 = 125h0 + 17h1 + 13h2 = m2
625h0 + 257h1 + 97h2 = m4
h0 + h1 + h2 = 1, (110)
and quadratic equations are
h0h0 + h1h1 + h2h2 = 1
h2h0 + h2h1 + (h0 + h1)h2 = 0
h1h0 + h0h1 = 0. (111)
In this system, there are six unknowns with four linear and three quadratic equations. Applying SINGULAR with lexicographicalordering and rankingh2 > h1 > h0 > h2 > h1 > h0 to the above sets of equations, the coefficients, in terms ofm2 andm4,
22
obtained from the reduced Grobner basis are
h0 = −40m2 −m4 − 423
48
h1 =44m2 −m4 − 475
16
h2 = −46m2 −m4 − 525
24
h0 =1840m2
2 − 86m2m4 − 39498m2 + m24 + 924m4 + 211923
15824m22 − 712m2m4 − 346488m2 + 8m2
4 + 7800m4 + 1896048
h1 =6072m2
2 − 270m2m4 − 131682m2 + 3m24 + 2928m4 + 713925
15824m22 − 712m2m4 − 346488m2 + 8m2
4 + 7800m4 + 1896048
h2 =46m2 −m4 − 525
92m2 − 2m4 − 1026. (112)
Using m2 and m2
Since there is only one reduced vanishing moment in both filters, the condition in Theorem 1 is not satisfied. Therefore, thediscrete momentm2 is still independent ofm2. The linear equations of analysis and synthesis filter coefficients are
h0 + h1 + h2 = 125h0 + 17h1 + 13h2 = m2
h0 + h1 + h2 = 1
25h0 + 17h1 + 13h2 = m2, (113)
and the quadratic equations are
h0h0 + h1h1 + h2h2 = 1
h2h0 + h2h1 + (h0 + h1)h2 = 0
h1h0 + h0h1 = 0. (114)
Applying SINGULAR with lexicographical ordering and rankingh2 > h1 > h0 > h2 > h1 > h0 to the above sets ofequations, the first polynomial equation , parametrized bym2 andm2, obtained from the reduced Grobner basis is a quadraticequation inh0:
(128m2 − 2048)h20 + (−24m2m2 + 376m2 + 344m2 − 5304)h0+
(m22m2 − 15m2
2 − 28m2m2 + 416m2 + 195m2 − 2873) = 0.
Solve the above equation forh0 and substitute into the following equations to get the other coefficients.
4h1 + 12h0 + (−m2 + 13) = 0
4h2 − 8h0 + (m2 − 17) = 0
(16m2 − 256)h0 + (16m2 − 256)h0 + (−3m2m2 + 45m2 + 45m2 − 663) = 0
(16m2 − 256)h1 + (−48m2 + 768)h0 + (5m2m2 − 83m2 − 71m2 + 1157) = 0
(8m2 − 128)h2 + (16m2 − 256)h0 + (−m2m2 + 11m2 + 13m2 − 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,V3, thereduced vanishing moments of analysis filter and synthesis filter can be (1,3) or (2,2) with one parameter. If one gives up onemore reduced vanishing momentV2 for two parameters, then the allocation of reduced vanishing moments will become (2,1).
23
Reduced Vanishing moments (1,3)
When there are three reduced vanishing moments inψ, the linear equations for the filter coefficients ofφ are
h0 + h1 + h2 + h3 = 17h0 − 5h1 + 3h2 − h3 = 0
91h0 − 35h1 + 9h2 − h3 = 049h0 + 37h1 + 29h2 + 25h3 = m2. (116)
After solving the preceding linear system, one has
h0 =m2 − 27
64
h1 =3m2 − 77
64
h2 =m2 − 7
64
h3 =−5m2 + 175
64. (117)
The equations for the dual filter coefficientsφ are
h0 + h1 + h2 + h3 =1
h0h0 + h1h1 + h2h2 + h3h3 =1
h2h0 + h3h1 + (h0 + h3)h2 + (h1 + h2)h3 =0
h3h0 + h2h1 + h1h2 + h0h3 =0. (118)
where the first one is from normalization and the first sum rule is eliminated by invoking the symmetry constraint on coefficients.Substitutehk in (117) into the preceding system of equations to get thehkcoefficients :
h0 =m3
2 − 78m22 + 2031m2 − 17658
32(m2 − 29)
h1 =−3m3
2 + 230m22 − 5889m2 + 50358
32(m2 − 29)
h2 =3m3
2 − 230m22 + 5905m2 − 50694
32(m2 − 29)
h3 =−m3
2 + 78m22 − 2015m2 + 17066
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 equationsfor analysis filter coefficients are
h0 + h1 + h2 + h3 = 17h0 − 5h1 + 3h2 − h3 = 0
49h0 + 37h1 + 29h2 + 25h3 = m2. (120)
and the linear equations for synthesis filter coefficients are
h0 + h1 + h2 + h3 = 1
7h0 − 5h1 + 3h2 − h3 = 0. (121)
Applying SINGULAR with lexicographical ordering and rankingh3 > ... > h0 > h3 > ... > h0 to the above sets of equationsand the following quadratic equations
h0h0 + h1h1 + h2h2 + h3h3 = 1
h2h0 + h3h1 + (h0 + h3)h2 + (h1 + h2)h3 = 0
h3h0 + h2h1 + h1h2 + h0h3 = 0, (122)
24
the first polynomial equation forh0 with coefficients parametrized bym2, obtained in the Gronber basis, is
2048h20 + (−32m2
2 + 1440m2 − 16192)h0 + (m32 − 75m2
2 + 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) = 016h2 + 48h0 + (−m2 + 22) = 0
8h3 − 24h0 + (m2 − 32) = 0
64h0 + 64h0 + (−m22 + 49m2 − 604) = 0
64h1 − 64h0 + (m22 − 45m2 + 512) = 0
64h2 − 192h0 + (3m22 − 143m2 + 1704) = 0
64h3 + 192h0 + (−3m22 + 139m2 − 1676) = 0. (124)
For the special case, whenm2 = 23 the h1 and h2 will be zeros and
hk = [332
, − 932
, − 732
,4532
,4532
, − 732
, − 932
,332
]
hk = [0, 0,14,
34,
34,
14, 0, 0],
which is the biorthogonal spline wavelet bio3.3. The length difference is again an even multiple by 2.
Reduced Vanishing moments (1,2) and two parameters
The parametrization of biorthogonal wavelets with two parameters requires one more degree of freedom which results bygiving up one more reduced vanishing moment. The distribution of reduced vanishing moments will then be (1,2). By Theorem1, m2 will be linear dependent onm2, that is
m2 + m2 = m21 = N2.
The only realization of the parametrization is with parametersm2 andm4.
Using m2 and m4
The linear equations forhk are
h0 + h1 + h2 + h3 = 17h0 − 5h1 + 3h2 − h3 = 0
49h0 + 37h1 + 29h2 + 25h3 = m2
24010 + 1297h1 + 641h2 + 337h3 = m4 (125)
and one linear equation for synthesis filter coefficientshk is
h0 + h1 + h2 + h3 = 1. (126)
The quadratic equations are
h0h0 + h1h1 + h2h2 + h3h3 = 1
h2h0 + h3h1 + (h0 + h3)h2 + (h1 + h2)h3 = 0
h3h0 + h2h1 + h1h2 + h0h3 = 0. (127)
Applying SINGULAR with lexicographical ordering and rankingh3 > ... > h0 > h3 > ... > h0 to the above sets of equations,the coefficients obtained in the Grobner basis, parametrized bym2 andm4 are
h0 =− 79m2 −m4 − 1641192
h1 =91m2 −m4 − 1953
192
h2 =83m2 −m4 − 1729
64
h3 =− 87m2 −m4 − 189764
(128)
25
and
h0 =6715m3
2 − 164m22m4 − 430126m2
2 + m2m24 + 7058m2m4 + 9185073m2 − 22m2
4 − 75948m4 − 65387286
−223040m22 + 5344m2m4 + 9422496m2 − 32m2
4 − 112896m4 − 99509472
h1 =7735m3
2 − 176m22m4 − 500794m2
2 + m2m24 + 7634m2m4 + 10811073m2 − 22m2
4 − 82812m4 − 77819238
−223040m22 + 5344m2m4 + 9422496m2 − 32m2
4 − 112896m4 − 99509472
h2 =7735m3
2 − 176m22m4 − 389274m2
2 + m2m24 + 4962m2m4 + 6131313m2 − 6m2
4 + 26748m4 − 28724598
223040m22 − 5344m2m4 − 9422496m2 + 32m2
4 + 112896m4 + 99509472
h3 =6715m3
2 − 164m22m4 − 318606m2
2 + m2m24 + 4386m2m4 + 4442337m2 − 6m2
4 − 19116m4 − 14972454
223040m22 − 5344m2m4 − 9422496m2 + 32m2
4 + 112896m4 + 99509472. (129)
F. 10/10 case
When the lengthl = 10, the maximum number of reduced vanishing moments is five. Give up the highest one to get oneparameterm2. The distribution of reduced vanishing moments can either be (1,4) or (2,3). If one gives up one more reducedvanishing moment to obtain two parametersm2 andm4, the reduced vanishing moments can be distributed as (1,3) or (2,2).Note, m2 will be dependent onm2 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 equationsare as follows,
h0 + h1 + h2 + h3 + h4 = 1−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0
−189h0 + 91h1 − 35h2 + 9h3 − h4 = 0−4149h0 + 1267h1 − 275h2 + 33h3 − h4 = 0
81h0 + 65h1 + 53h2 + 45h3 + 41h4 = m2. (130)
From the above, directly get thehk coefficients.
h0 =m2 − 44
256
h1 =5m2 − 216
256
h2 =2m2 − 81
64
h3 =2164
h4 =−7m2 + 378
128(131)
Substitute the abovehk into following linear and quadratic equations forhk
h0 + h1 + h2 + h3 + h4 = 1
h0h0 + h1h1 + h2h2 + h3h3 + h4h4 = 1
h2h0 + h3h1 + (h0 + h4)h2 + (h1 + h4)h3 + (h2 + h3)h4 = 0
h4h0 + h4h1 + h3h2 + h2h3 + (h0 + h1)h4 = 0
h3h0 + h2h1 + h1h2 + h0h3 = 0
h1h0 + h0h1 = 0, (132)
26
to obtain thehk coefficients below.
h0 =−m4
2 + 172m32 − 11098m2
2 + 318364m2 − 3425840128(m2 − 46)
h1 =5m4
2 − 856m32 + 54978m2
2 − 1569956m2 + 16817760128(m2 − 46)
h2 =−5m4
2 + 854m32 − 54730m2
2 + 1559696m2 − 1667628064(m2 − 46)
h3 =5m4
2 − 854m32 + 54730m2
2 − 1559664m2 + 1667532064(m2 − 46)
h4 =−m4
2 + 171m32 − 10970m2
2 + 312914m2 − 3348972)32(m2 − 46)
(133)
Reduced Vanishing Moments (2,3)
When two reduced vanishing moments are allocated to wavelet functionψ and three to its dualψ, the linear system
h0 + h1 + h2 + h3 + h4 = 1−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0
−189h0 + 91h1 − 35h2 + 9h3 − h4 = 081h0 + 65h1 + 53h2 + 45h3 + 41h4 = m2
h0 + h1 + h2 + h3 + h4 = 1
−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0,
contains second and third sum rules forhk and second sum rule forhk. The quadratic equations are the same as in the previouscase (132). Applying SINGULAR with lexicographical ordering and rankingh4 > ... > h0 > h4 > ... > h0 to the above setsof equations and (132), the first polynomial equation forh0 with coefficients parametrized bym2, obtained in the Grobnerbasis, is
(−65536m2 + 2621440)h20 + (128m3
2 − 15232m22 + 603264m2 − 7948160)h0+
(−m42 + 168m3
2 − 10588m22 + 296680m2 − 3118403) = 0. (134)
Solve the above equation forh0 and then substitute into the following equations obtained from the same Grobner basis to getthe other coefficients.
h1 = h0 +m2 − 43
64
h2 = −4h0 +3m2 − 125
64
h3 = −4h0 +m2 − 23
64
h4 = 6h0 +−5m2 + 255
64
h0 = 2(m2 − 41)h0 − m32 − 125m2
2 + 5213m2 − 72521
256
h1 = −2(3m2 − 121)h0 +3m3
2 − 371m22 + 15311m2 − 210815
256
h2 = 4h0 − m22 − 78m2 + 1531
64
h3 = 4(4m2 − 161)h0 − 2m32 − 247m2
2 + 10184m2 − 140153
64
h4 = −12(m2 − 40)h0 +3m3
2 − 369m22 + 15163m2 − 207969
128. (135)
27
Reduced Vanishing Moments (1,3) and two Parameters
The second parameterm4 is obtained by giving up the fourth reduced vanishing momentV3. The linear system becomes
h0 + h1 + h2 + h3 + h4 = 1−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0
−189h0 + 91h1 − 35h2 + 9h3 − h4 = 081h0 + 65h1 + 53h2 + 45h3 + 41h4 = m2
6561h0 + 4097h1 + 2417h2 + 1377h3 + 881h4 = m4
h0 + h1 + h2 + h3 + h4 = 1. (136)
The quadratic equations are still the same:
h0h0 + h1h1 + h2h2 + h3h3 + h4h4 = 1
h2h0 + h3h1 + (h0 + h4)h2 + (h1 + h4)h3 + (h2 + h3)h4 = 0
h4h0 + h4h1 + h3h2 + h2h3 + (h0 + h1)h4 = 0
h3h0 + h2h1 + h1h2 + h0h3 = 0.
Applying SINGULAR with lexicographical ordering and rankingh4 > ... > h0 > h4 > ... > h0 to the above sets of equations,the coefficientshi and hi obtained in the Grobner basis, parametrized bym2 andm4, are
h0 = −130m2 −m4 − 4458
768
h1 = −118m2 −m4 − 3942
768
h2 =139m2 −m4 − 4833
192
h3 =133m2 −m4 − 4527
192
h4 = −140m2 −m4 − 4968
128(137)
and
(−6945792m22 + 103296m2m4 + 481019904m2 − 384m2
4 − 3576960m4 − 8327937024)h0+
(53040m42 − 798m3
2m4 − 5660624m32 + 3m2
2m24 + 39936m2
2m4 + 209895696m22+
153m2m24 + 52038m2m4 − 2982370608m2 −m3
4 − 8694m24 − 16558488m4 + 10367524800) = 0
(−6945792m22 + 103296m2m4 + 481019904m2 − 384m2
4 − 3576960m4 − 8327937024)h1+
(−48144m42 + 762m3
2m4 + 5095472m32 − 3m2
2m24 − 38664m2
2m4 − 187432752m22−
141m2m24 + 10662m2m4 + 2644256592m2 + m3
4 + 8178m24 + 14372712m4 − 9167515200) = 0
(−3472896m22 + 51648m2m4 + 240509952m2 − 192m2
4 − 1788480m4 − 4163968512)h2+
(−60384m42 + 852m3
2m4 + 6074240m32 − 3m2
2m24 − 35388m2
2m4 − 197030112m22−
195m2m24 − 614976m2m4 + 1826316864m2 + m3
4 + 10380m24 + 28334736m4 + 7621879680) = 0
(−3472896m22 + 51648m2m4 + 240509952m2 − 192m2
4 − 1788480m4 − 4163968512)h3+
(60384m42 − 852m3
2m4 − 6074240m32 + 3m2
2m24 + 35388m2
2m4 + 198766560m22+
195m2m24 + 589152m2m4 − 1945958976m2 −m3
4 − 10284m24 − 27445104m4 − 5561032320) = 0
(578816m22 − 8608m2m4 − 40084992m2 + 32m2
4 + 298080m4 + 693994752)h4+
(408m42 − 3m3
2m4 − 47096m32 + 106m2
2m4 + 1582504m22 + m2m2
4+
9529m2m4 − 8031528m2 − 59m24 − 331956m4 − 250519392) = 0.
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 = 1−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0
81h0 + 65h1 + 53h2 + 45h3 + 41h4 = m2
6561h0 + 4097h1 + 2417h2 + 1377h3 + 881h4 = m4
and
h0 + h1 + h2 + h3 + h4 = 1
−9h0 + 7h1 − 5h2 + 3h3 − h4 = 0.
The quadratic equations remain the same. Applying SINGULAR with lexicographical ordering and rankingh4 > ... > h0 >h4 > ... > h0 to the sets of equations, the first polynomial equation forh0 with coefficients parametrized bym2 and m4,obtained in the Grobner basis, is
(−884736m22 + 31555584m2 + 294912m4 − 79626240)h2
0+
(−452736m32 + 3456m2
2m4 + 31863168m22 + 26496m2m4 − 606450816m2 − 1152m2
4 − 4836096m4+
1585090944)h0 + (−50673m42 + 780m3
2m4 + 5335214m32 − 3m2
2m24 − 38475m2
2m4 − 193582242m22−
150m2m24 − 78588m2m4 + 2628672426m2 + m3
4 + 8541m24 + 16484391m4 − 7572966453) = 0
Solve the above equation forh0 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
(−34816m2 + 256m4 + 1209600)h0+
(768m22 − 27392m2 − 256m4 + 69120)h0+
(399m32 − 3m2
2m4 − 28409m22 − 23m2m4 + 555903m2 + m2
4 + 4212m4 − 1749357) = 0
(−104448m2 + 768m4 + 3628800)h1 + (−6912m22 + 246528m2 + 2304m4 − 622080)h0+
(−1959m32 + 15m2
2m4 + 135877m22 + 127m2m4 − 2482695m2 − 5m2
4 − 21420m4 + 4347513) = 0
192h2 + (3m22 − 104m2 −m4 + 135) = 0
(−8704m2 + 64m4 + 302400)h3+
(1536m22 − 54784m2 − 512m4 + 138240)h0+
(390m32 − 3m2
2m4 − 27411m22 − 22m2m4 + 523990m2 + m2
4 + 4130m4 − 1462239) = 0
(−17408m2 + 128m4 + 604800)h4+
(−2304m22 + 82176m2 + 768m4 − 207360)h0+
(−381m32 + 3m2
2m4 + 27501m22 + 13m2m4 − 554901m2 −m2
4 − 3864m4 + 2044521) = 0.
Length Odd Case
If the filter length for scaling function is odd, then the symmetry constrainthk = hN−k transforms the normalizationequation forφ to
2
N2 −1∑
k=0
hk + hN/2 = 2, (138)
29
For φ the normalization equation becomes
2
N2 −1∑
k=0
hk + hN/2 = 2. (139)
Furthermore, the equation from sum rule becomes
n−1∑
k=0
(−1)n−k+1[(n− k + 1)l + (k + 1− n)l]hk − hn = 0,
whereN = 2n and l = 0, 1, ..., p− 1, p is vanishing moments ofψ.With the symmetric coefficients, the discrete moments will become
mn =N∑
k=0
hkkn =
N2 −1∑
k=0
(kn + (N − k)n)hk + (N
2)nhN/2,
where N is even andn is the order of discrete moment. The odd-indexed moments are linearly dependent on even-indexedmoments, described in equation (72) to (74). Sincem0 andm1 are constant,m2 will be chosen as parameter. In the following,when the length is five, the maximum number of reduced vanishing moments isp = (5 − 1)/2 = 2 and the the number ofdegree of freedom to allocate reduced vanishing moments isp − 2 = 0, which means that length 5/5 case has no degree offreedom for parametrization.
G. 5/5 case
From above the only possible allocation in (ψ, ψ) is (1,1) for this case. The linear equations obtained from normalization,sum rules and discrete moment are
2h0 + 2h1 + h2 = 2−2h0 + 2h1 − h2 = 0
16h0 + 10h1 + 4h2 = m2
2h0 + 2h1 + h2 = 2
−2h0 + 2h1 − h2 = 0. (140)
The quadratic equations are
2h0h0 + 2h1h1 + h2h2 = 2
h0h2 + h1h1 + h2h0 = 0. (141)
Solve the above sets of equations to getm2 = 7 or m2 = 9. For m2 = 7, the corresponding filter coefficients are
h = [−0.25 0.5 1.5 0.5 − 0.25]
h = [0 0.5 1 0.5 0]. (142)
When m2 = 9, the two filter coefficients are the same but just interchanged. This is exactly the case of biorthogonal splinewavelet bio 2.2 [9].
H. 7/7 caseReduced Vanishing Moment (1,1)
In this case, the maximum number of reduced vanishing moments will be three. Give up one reduced vanishing momentto obtain one degree of freedom for parametrization. The remaining number of reduced vanishing moments is two and thenumber of degree of freedom for allocation is 0. Therefore, the only way to allocate the moment is (1,1) in(φ, φ). The linearsystem becomes:
2h0 + 2h1 + 2h2 + h3 = 22h0 − 2h1 + 2h2 − h3 = 0
36h0 + 26h1 + 20h2 + 9h3 = m2
2h0 + 2h1 + 2h2 + h3 = 2
2h0 − 2h1 + 2h2 − h3 = 0. (143)
30
The quadratic equations are
2h0h0 + 2h1h1 + 2h2h2 + h3h3 = 2
h2h0 + h3h1 + (h0 + h2)h2 + h1h3 = 0
h0h2 + h1h1 + h2h0 = 0. (144)
Applying SINGULAR with lexicographical ordering and rankingh3 > ... > h0 > h3 > ... > h0 to the above sets of equations,the first polynomial equation forh0 with coefficients parametrized bym2 and the other polynomial equations obtained in theGrobner basis are
(−16m2 + 240)h20 + (m2
2 − 36m2 + 323)h0 = 08h1 + 16h0 + (−m2 + 19) = 0
2h2 + 2h0 − 1 = 04h3 − 16h0 + (m2 − 23) = 0
(−16m2 + 336)h0 + (16m2 − 240)h0 + (−m22 + 36m2 − 323) = 0
(−4m2 + 84)h1 + (−8m2 + 120)h0 + (m2 − 17) = 0
(−16m2 + 336)h2 + (−16m2 + 240)h0 + (m22 − 28m2 + 155) = 0
(−2m2 + 42)h3 + (8m2 − 120)h0 + (m2 − 25) = 0. (145)
Usefacstd command in SINGULAR to factorize the above equations and get, in the first part,
h0 = 0, h1 =m2 − 19
8, h2 =
1
2, h3 =
−m2 + 23
4
h0 =m2
2 − 36m2 + 323
−16m2 + 336, h1 =
m2 − 17
4m2 − 84, h2 =
m22 − 28m2 + 155
16m2 − 336, h3 =
m2 − 25
2m2 − 42. (146)
and, in the second part,
h0 =m2
2 − 36m2 + 32316m2 − 240
, h1 =m2 − 194m2 − 60
, h2 =m2
2 − 44m2 + 443−16m2 + 240
, h3 =m2 − 112m2 − 30
h0 = 0, h1 =−m2 + 17
8, h2 =
12, h3 =
m2 − 134
. (147)
Note that the preceding two cases correspond to the length 7/5 and 5/7 cases. For the special case, add one more reducedvanishing moment onψ to get one more sum rule forhk, that is
20h0 − 10h1 + 4h2 − h3 = 0. (148)
Substitute the above parameterized coefficients into this sum rule to getm2 = 20 and
h0 = 0, h1 =18, h2 =
12, h3 =
34,
h0 =316
, h1 =−34
, h2 =516
, h3 =52. (149)
I. 9/9 case
When the length is longer, there are more possibilities for allocating the reduced vanishing moments. The maximum numberof 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, thenumber of degree freedom to allocate reduced vanishing moments isp − 2 = 0. When the the number of reduced vanishingmoments is maximum, one has the special cases of (3,1) and (2,2) moments in(φ, φ). These different cases are discussed inthe following subsections.
31
Reduced vanishing moments (1,2)
When there are two reduced vanishing moments inψ there are two sum rules forhk. The linear equations are
2h0 + 2h1 + 2h2 + 2h3 + h4 = 22h0 − 2h1 + 2h2 − 2h3 + h4 = 0
−34h0 + 20h1 − 10h2 + 4h3 − h4 = 064h0 + 50h1 + 40h2 + 34h3 + 16h4 = m2
2h0 + 2h1 + 2h2 + 2h3 + h4 = 2
2h0 − 2h1 + 2h2 − 2h3 + h4 = 0. (150)
The quadratic equations are
2h0h0 + 2h1h1 + 2h2h2 + 2h3h3 + h4h4 = 2
h2h0 + h3h1 + (h0 + h4)h2 + (h1 + h3)h3 + h2h4 = 0
h4h0 + h3h1 + h2h2 + h1h3 + h0h4 = 0
h2h0 + h1h1 + h0h2 = 0. (151)
Applying SINGULAR with lexicographical ordering and rankingh4 > ... > h0 > h4 > ... > h0 to the above sets of equationsand factorizing the polynomial equations obtained in the Grobner basis, the first set of coefficients is
h0 = 0, h1 =m2 − 34
32, h2 =
m2 − 32
16, h3 = −m2 − 50
32, h4 = −m2 − 40
8
h0 =m3
2 − 99m22 + 3270m2 − 36040
64m2 − 2304, h1 =
−m32 + 97m2
2 − 3140m2 + 33920
32m2 − 1152, h2 =
m2 − 28
4m2 − 144,
h3 =m3
2 − 97m22 + 3156m2 − 34496
32m2 − 1152, h4 =
−m32 + 99m2
2 − 3254m2 + 35336
32m2 − 1152. (152)
and the other set is
8192h20 + (−64m2
2 + 3904m2 − 59520)h0 + (m32 − 99m2
2 + 3270m2 − 36040) = 0
32h1 + (−m2 + 34) = 0
16h2 + 64h0 + (−m2 + 32) = 0
32h3 + (m2 − 50) = 0
8h4 − 48h0 + (m2 − 40) = 0
h0 = 0
32h1 + 128h0 + (−m22 + 65m2 − 1060) = 0
16h2 − 128h0 + (m22 − 63m2 + 998) = 0
32h3 − 128h0 + (m22 − 65m2 + 1044) = 0
8h4 + 128h0 + (−m22 + 63m2 − 1006) = 0. (153)
For the second factorized parametrization, first solve the equation forh0 and them substitute into the other equations to getthe coefficients.
Reduced vanishing moments (1,1) and two Parameters
If one more reduced vanishing moments in analysis filter is given up, the fourth discrete momentm4 can be used as thesecond parameter. The linear equations are
2h0 + 2h1 + 2h2 + 2h3 + h4 = 22h0 − 2h1 + 2h2 − 2h3 + h4 = 0
64h0 + 50h1 + 40h2 + 34h3 + 16h4 = m2
4096h0 + 2402h1 + 1312h2 + 706h3 + 256h4 = m4
2h0 + 2h1 + 2h2 + 2h3 + h4 = 2
2h0 − 2h1 + 2h2 − 2h3 + h4 = 0. (154)
and the quadratic equations are still the same. Applying SINGULAR with lexicographical ordering and rankingh4 > ... >h0 > h4 > ... > h0 to the above sets of equations and factorizing the polynomial equations obtained in the Grobner basis, the
32
first set of coefficients is
h0 = 0
h1 = −100m2 −m4 − 2691
96
h2 =106m2 −m4 − 2889
48
h3 =100m2 −m4 − 2643
96
h4 = −106m2 −m4 − 2913
24
h0 = −10600m32 − 206m2
2m4 − 882746m22 + m2m2
4 + 11566m2m4 + 24496425m2 − 29m24 − 162216m4 − 226520307
698752m22 − 13376m2m4 − 38050752m2 + 64m2
4 + 364224m4 + 518002560
h1 = −11236m32 − 212m2
2m4 − 939584m22 + m2m2
4 + 11938m2m4 + 26184537m2 − 29m24 − 167958m4 − 243187353
349376m22 − 6688m2m4 − 19025376m2 + 32m2
4 + 182112m4 + 259001280
h2 =106m2 −m4 − 2877
424m2 − 4m4 − 11604
h3 =11236m3
2 − 212m22m4 − 764896m2
2 + m2m24 + 8594m2m4 + 16671849m2 − 13m2
4 − 76902m4 − 113686713
349376m22 − 6688m2m4 − 19025376m2 + 32m2
4 + 182112m4 + 259001280
h4 =10600m3
2 − 206m22m4 − 708058m2
2 + m2m24 + 8222m2m4 + 14944185m2 − 13m2
4 − 70776m4 − 95948307
349376m22 − 6688m2m4 − 19025376m2 + 32m2
4 + 182112m4 + 259001280(155)
The second set contains a quadratic equation inh0 with coefficients parametrized bym2 andm4, i.e.
(−73728m2 + 2064384)h20 + (−59904m2
2 + 576m2m4 + 3327168m2 − 16320m4 − 46202688)h0 − (10600m32−
206m22m4 − 882746m2
2 + m2m24 + 11566m2m4 + 24496425m2 − 29m2
4 − 162216m4 − 226520307) = 0. (156)
After solving this equation, substitute the solution forh0 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
h0 = 0
(3488m2 − 32m4 − 96192)h1 + (384m2 − 10752)h0 + (318m22 − 3m2m4 − 17925m2 + 87m4 + 252531) = 0
(−1744m2 + 16m4 + 48096)h2 + (384m2 − 10752)h0 + (100m22 −m2m4 − 5155m2 + 25m4 + 66159) = 0
(3488m2 − 32m4 − 96192)h3 + (−384m2 + 10752)h0 + (−318m22 + 3m2m4 + 16181m2 − 71m4 − 204435) = 0
(−872m2 + 8m4 + 24048)h4 + (−384m2 + 10752)h0 + (−100m22 + m2m4 + 6027m2 − 33m4 − 90207) = 0 (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 isto equally distribute on both sides and the other is to put more on one side. The smoothness of wavelet function will dependon the number of vanishing moments allocated to it. In the equally distributed case, that is (2,2), one has the following linearequations
2h0 + 2h1 + 2h2 + 2h3 + h4 = 22h0 − 2h1 + 2h2 − 2h3 + h4 = 0
−34h0 + 20h1 − 10h2 + 4h3 − h4 = 0
2h0 + 2h1 + 2h2 + 2h3 + h4 = 2
2h0 − 2h1 + 2h2 − 2h3 + h4 = 0
−34h0 + 20h1 − 10h2 + 4h3 − h4 = 0, (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
h0 = 0, h1 = −0.0912, h2 = −0.0576, h3 = 0.5912, h4 = 1.1150. (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 oflength 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 Daubechieswavelet db4.
which is biorthogonal spline wavelet bio4.4. [15] In the (1,3) case, there are three sum rules forhk and one forhk. The linearequations become
2h0 + 2h1 + 2h2 + 2h3 + h4 = 22h0 − 2h1 + 2h2 − 2h3 + h4 = 0
−34h0 + 20h1 − 10h2 + 4h3 − h4 = 0−706h0 + 272h1 − 82h2 + 17h3 − h4 = 0
2h0 + 2h1 + 2h2 + 2h3 + h4 = 2
2h0 − 2h1 + 2h2 − 2h3 + h4 = 0, (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
h0 = 0, h1 = 0.2015, h2 = −0.5535, h3 = 0.2985, h4 = 2.1070, (161)
and
h0 = 0, h1 = 0.0263, h2 = 0.1776, h3 = 0.4737, h4 = 0.6447
h0 = −0.1049, h1 = 0.7084, h2 = −1.4773, h3 = −0.2084, h4 = 4.1644. (162)
IV. EXAMPLES
The choice of mother wavelet is application-dependent [16]. Parametrization with two parameters offers more choicesthan parametrization with one parameter. In [1], an example on signal compression is given within the setting of waveletdecomposition level four. The square error and maximum error of the ’optimal’ wavelets Regensburger proposed is better thanDaubechies wavelet (db4). Here, the same input signal with the same setting are tested and one can see the wavelet proposedhere 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. CONCLUSIONS
Filter bank parametrization is useful in performance evaluation with a large number of different wavelets, in a convenientmanner. Recently, Regensburger and Scherzer [1] used one discrete moment as parameter in the parametrization of FIR filtercoefficients to generate orthogonal wavelets after establishing a bijective relationship between continuous and discrete momentsof 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 bankcan be used. This is more complicated because the problem of generating discrete scaling functions and mother wavelets whichdiffer in length may be encountered. A characterization theorem for continuous moments of the scaling function and its dualwas formulated and proved for any biorthogonal system. Regensburger’s result emerged as a special case of this general resultgiven here as Theorem 1.
The possibility of using Grobner bases in the parametrization of filter bank coefficients for orthogonal wavelets was consideredand shown to lead to better performance evaluation in a generic image compression example that was used by Regensburger andScherzer [1] to show the superiority of his parametrization approach with one discrete moment over the classical Daubechiesorthogonal system. It is shown that Grobner bases are effective in formulating and solving a parametrized system of filtercoefficients used for generating biorthogonal wavelets. The important special case of symmetric biorthogonal wavelets arediscussed in detail for FIR filters of different lengths. The systematic generation of filters provide a mechanism for generatinga wide variety of wavelets including many belonging to distinguished classes for selection of optimum application-dependentmother wavelets [16].
APPENDIX
Length Eight Orthogonal Case: The equations left in the triangular system mentioned in length eight case are:
(13824m41 − 163968m3
1 − 5376m21m3 + 725888m2
1 + 32768m1m3 − 1381632m1 + 512m23−
53760m3 + 898560)h1 − 2359296h30 + (−405504m2
1 + 1671168m1 + 98304m3 − 958464)h20+
(−14400m41 + 82176m3
1 + 7680m21m3 + 32768m2
1 − 22528m1m3 − 686592m1 − 1024m23 − 27648m3+
707328)h0 + (117m61 − 2802m5
1 − 60m41m3 + 24296m4
1 + 1184m31m3 − 99320m3
1 + 8m21m
23−
7184m21m3 + 206960m2
1 − 128m1m23 + 14912m1m3 − 217320m1 + 464m2
3 − 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 + 1013760m6
1 + 36864m51m3 − 4810752m5
1 − 331776m41m3 − 62951424m4
1−5406720m3
1m3 + 633249792m31 − 98304m2
1m23 + 72990720m2
1m3 − 2228699136m21+
1769472m1m23 − 257064960m1m3 + 3460644864m1 − 7864320m2
3 + 307003392m3 − 1893556224)h1+
7247757312h30 + (14155776m4
1 − 2868903936m21 − 75497472m1m3 + 16590569472m1+
792723456m3 − 31255953408)h20 + (6912m8
1 − 3022848m61 − 73728m5
1m3 + 17879040m51+
811008m41m3 + 297762816m4
1 + 16121856m31m3 − 3953049600m3
1 + 196608m21m
23−
272990208m21m3 + 19472203776m2
1 − 4325376m1m23 + 1238237184m1m3 − 45856751616m1+
23789568m23 − 2047279104m3 + 44454887424)h0 − (108m10
1 − 765m91 − 18m8
1m3 − 47448m81−
1152m71m3 + 657264m7
1 + 28992m61m3 + 2204592m6
1 + 192m51m
23 + 103488m5
1m3 − 106356640m51+
960m41m
23 − 7099200m4
1m3 + 1010070976m41 − 132864m3
1m23 + 70831104m3
1m3 − 5047712000m31−
512m21m
33 + 1659648m2
1m23 − 331249920m2
1m3 + 14768421504m21 + 11264m1m
33 − 7317504m1m
23+
784624128m1m3 − 24221738304m1 − 62464m33 + 12056064m2
3 − 780107904m3 + 17484139008) = 0
1536h2 + 3072h1 + 9216h0 + (3m41 − 836m2
1 − 16m1m3 + 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
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