PARAMETERIZATION OF MULTIDIMENSIONAL DECIMATION MATRICES M.K. Tchobanou
Moscow Power Engineering Institute (Technical University) Department of Electrical Physics 13 Krasnokazarmennaya st., 105835, Moscow, RUSSIA Tel.: +7(095)362-7463,
[email protected] ABSTRACT A general method for multidimensional nonseparable decimation matrices design was developed. It allows to fully parameterize all the family of such decimation matrices. The main attention was paid to finding of the decimation matrix eigenvalues. 1. MULTI-DIMENSIONAL MULTIRATE SYSTEMS The growing demand for processing and compression of still two-dimensional (2-D) images and video (3-D) signals in telecommunications and multimedia technology motivates the fact that increasingly more attention is being paid to multi-dimensional (M-D) digital systems. As usual the transformation part of the encoding system is performed by some multirate systems. The main interest in this paper is connected with so-called true multidimensional systems, or nonseparable multirate systems. The decimation/interpolation processes are implemented by use of some tools. One of the most important is the decimation matrix. 2. REQUIREMENTS FOR DECIMATION MATRICES Nonseparable decimation matrices and FBs are preferable because M-D signals by their nature are nonseparable. Nonseparable FBs have better characteristics than their separable counterparts (which consist of products of 1-D FBs along each dimension). The number of degrees of freedom is also much bigger for nonseparable FBs. There was not any systematic approach that allowed to build nonseparable decimation matrices for given dimension and given number of channels. In what follows such an approach is given. The main requirements that should be met by decimation matrices are the next: 1. The matrix elements are integer numbers. 2. The number of the channels is given by the absolute value of the decimation matrix determinant - m = |det(M )|.
3. Towards to be sure that in each direction the dilation takes place it is necessary to have the eigenvalues of the decimation matrix V to be bigger than 1 [1]. 4. From implementation point of view it is also necessary to achieve separable decimation after a few steps of iterations. It means that the decimation matrix M should be n-th square of identity matrix: M n = c · Ik for a small value n, k- the number of dimensions, I- identity matrix. 3. 2-D CASE The decimation matrix equals ¸ · a b . M= c d
(1)
The eigenvalues might be found from det(M − λI) = 0. Thus we get the next system of equations: ad − bc = ±N λ2 − λ(a + d) + ad − bc = 0 . |λ| > 1
(2)
(3)
Bezout theorem states that the roots of the square equation should satisfy ½ λ1 + λ2 = a + d . (4) λ1 λ2 = ad − bc
It means that the sum and the product of these two roots has to be integer and real. If to use the polar representation for complex numbers λ1 and λ2 then we get λ1 = ρ1 exp (iφ1 ), λ2 = ρ2 exp (iφ2 ).
(5)
After substitution it into (4) one gets the system of equations ρ1 ρ2 = N, φ1 + φ2 = 0, if det(M ) = +N φ1 + φ2 = π, if det(M ) = −N (6) ρ sin φ + ρ sin φ = 0, 1 1 2 2 ρ1 cos φ1 + ρ2 cos φ2 = L
nonzero solutions exist for N > 2. If N =√3 then we have three solutions: L =√0, ρ1 = ρ2 = 3 and √ 13±1 L = ±1, ρ1 = 2 , ρ2 = 13∓1 . 2
where L = a + d is an integer number. Positive determinant det(M ) = +N . It means that φ1 + φ2 = 0. Then the solutions are: 1.
2. φ1 = 0 φ =0 2 ρ + ρN1 = L 1 ρ2 = ρN1 .
Remind that ρk > 1 and correspondingly ρk < N . The p solution for this equation would be ρ1 = L/2± L2 /4 − N for positive integers L. This equation has solution√(with the restrictions given above for ρk ) when 2 N < L < N + 1 or when N ≥ 5. √
√
If N = 5 then L = 5 and ρ1 = 5+2 5 , ρ1 = 5−2 5 . If N = 4 then ρ1 = ρ2 = 2 and therefore L = 4; if N = 6 then ρ1 = 2, ρ2 = 3 or ρ1 = 2, ρ2 = 3 and therefore L = 5 and so on. 2.
φ1 = π φ = −π 2 ρ + ρN1 = −L 1 ρ2 = ρN1 .
The same results as in the previous case, just the value of L changes the sign. 3.
L cos φ1 = 2√N φ = −φ1 √ 2 ρ1 = ρ2 = N
The solutions would be 0, ±1, ±2, L= ..., ±b2√N c,
φ1 = π2 cos φ1 = cos φ1 = ... cos φ1 =
φ1 = π φ =0 2 ρ − ρN1 = −L 1 ρ2 = ρN1 .
The solutions are the same as for previous case. Thus all possible values of the eigenvalues for 2-D case were found. At the same time there were found all possible values of L for the sum a + d. That is why one can write two equations with four integer unknowns: ½ a+d=L (7) ad − bc = ±N Depending on the sign and on the value of N there should be chosen proper values for L (see above). For example when determinant is −2 then L = 0 and therefore a = −d. Then one should solve the equation in integer unknowns −a2 − bc = −2 or a2 + bc = · 2. ¸ 0 1 If a = 0 we get the decimation matrices , 2 0 ¸ ¸ · ¸ · · 0 −2 0 −1 0 2 . If a = 1 we and , −1 0 −2 0 1 0 · ¸ · ¸ 1 1 1 −1 get the quincunx matrix and . 1 ¸ −1 · −1 ¸ −1 · −1 −1 −1 1 and so and If a = −1 we get −1 1 1 1 on. 3.1. M n = c · I2
1 √ 2 N 2 √ 2 N √ b2 N c √ , 2 N
where bxc is the floor function of x. For N = 2 one gets L = 0, ±1, ±2; for N = 3 we get L = 0, ±1, ±2, ±3 and so on.
In order to meet the requirement M·n = c·I2 one could no-¸ a2 + bc b(a + d) . tice that for n = 2 we have M 2 = c(a + d) d2 + bc That is why the second power of V would be at least diagonal if L = a + d = 0 (otherwise b = c = 0 and we get the separable case of decimation matrix which is of no interest). For positive determinant case there would be only one solution: √ π φ1 = −φ2 = , ρ1 = ρ2 = N . (8) 2 For negative determinant case we get
Negative determinant det(M ) = −N . It means that φ1 + φ2 = π. Then the solutions are: 1.
φ1 = 0 φ =π 2 ρ − ρN1 = L 1 ρ2 = ρN1 . √ The solution ρ1 = ρ2 = N is always available for a + d = 0. p Another solution is ρ1 = L/2 ± L2 /4 + N for integers L according to 1 − N < L < N − 1. The
φ1 = 0, φ2 = π, ρ1 = ρ2 =
√
N.
(9)
Similar results are obtained for n = 3, 4, . . .. 3.2. Eigenvalues On another and it is known that if λ1 , . . . , λk are the eigenvalues for the matrix M and g(u) is a scalar polynomial, then g(λ1 ), . . . , g(λk ) will be the eigenvalues for g(V ) [2]. That is why the n-th power of M is identity matrix multiplied by a scalar iff λn1 = . . . = λnk = c. 1 Therefore ρ1 = . . . = ρk = |c| n and nφ1 = . . . = nφk = arg c ( mod 2π). For n = 2 and k = 2 we have |c| = N this is exactly the same as (8) and (9).
4. 3D CASE The decimation matrix equals a b M = d e g h
c f . k
(10)
The eigenvalues are to be found from (2). Thus we get λ3 −(λ1 +λ2 +λ3 )λ2 +(λ1 λ2 +λ1 λ3 +λ2 λ3 )λ−λ1 λ2 λ3 = 0 (11) or λ3 − (λ1 + λ2 + λ3 )λ2 + λ1 λ2 λ3 ( λ11 + λ12 + λ13 )λ − λ1 λ2 λ3 = 0 From Bezout theorem we get λ1 + λ2 + λ3 = L λ1 λ2 + λ1 λ3 + λ2 λ3 = λ1 λ2 λ3 ( λ11 + λ1 λ2 λ3 = ±N
1 λ2
+
(12)
1 λ3 )
=K ,
(13) where L = a + e + k, K = ak − gc − db + ek + ae − f h, N = aek − af h + dhc − dbk + gbf − gec, L, K, N are integer numbers. If to use the polar representation (5) then after substitution into (13) we get the next system of equations:
amazing that there exist only seven distinct solutions and corresponding values for L and K (ρ2 = ρ3 , φ1 = 0 and φ3 = −φ2 ): 1. L =r0, K = −2 p √ 3 1 ρ2 = 1/3 3 46 + 6 57 + 12 √ −6= √ 3 46+6
44+3
Positive determinant det(M ) = +N . The solutions of this system of equations are the next: φ1 = 0 φ3 = −φ2 ρ3 = ρ2 N 1. ρ1 = ρ22 N + 2ρ2 cos φ2 = L ρ2 ρ22 + 2 cos φ N 1 = K 2 2 ρ2 φ1 = φ2 = φ3 = 0 ρ = N 1 ρ2 ρ3 2. N ρ + ρ 2 3 + ρ2 ρ3 = L 1 ρ2 ρ3 + N ( ρ2 + ρ13 ) = K
It is obvious that the first solution might be circularly changed to φ2 = 0, φ1 = −φ3 , ρ1 = ρ3 . . . and so on. We will consider only the variant mentioned in the first solution. For N = 2 a thorough analysis gives the values for the integers L and K that allow an admissible solution. It is
177
= 1.288008960, ρ1 = 1.20557, φ2 = 2.598652. 3. L =r0, K = −1 p √ 3 1 −3= ρ2 = 1/3 3 53 + 6 78 + 3 √ √ 3 53+6
78
= 1.146558421, ρ1 = 1.52138, φ2 = 2.296223 . 4. L = K = 0 ρ2 = 21/3 = 1.259921050, ρ1 = 1.259921050, φ2 = 2π/3. 5. L =r1, K = 0 p √ 3 1 = ρ2 = 1/3 3 54 + 6 87 − 18 √ √ 3 54+6
87
= 1.086052036, ρ1 = 1.695620, φ2 = 1.896792. 6. L =rK = 1 p √ 3 1 ρ2 = 1/3 3 46 + 3 249 − 15 √ +3= √ 3 46+3
249
= 1.215716761, ρ1 = 1.353211, φ2 = 1.71658. 7. L =rK = 2 p √ 3 1 ρ2 = 1/3 3 26 + 6 33 − 24 √ +6= √ 3 26+6
ρ1 ρ2 ρ3 = N, φ1 + φ2 + φ3 = 0, if det(M ) = +N φ1 + φ2 + φ3 = π, if det(M ) = −N ρ1 sin φ1 + ρ2 sin φ2 + ρ3 sin φ3 = 0, . 1 1 1 sin φ + sin φ + sin φ = 0, 1 2 3 ρ1 ρ2 ρ3 ρ1 cos φ1 + ρ2 cos φ2 + ρ3 cos φ3 = L ±N ( ρ11 cos φ1 + ρ12 cos φ2 + ρ13 cos φ3 ) = K
57
= 1.063200562, ρ1 = 1.7693, φ2 = 2.553606. 2. L =rK = −1 p √ 3 1 ρ2 = 1/3 3 44 + 3 177 + 21 √ −3= √ 3
33
= 1.138243270, ρ1 = 1.54369, φ2 = 1.368983. The second system does not have any solution satisfying all requirements for λk .
Negative determinant det(M ) = −N . The solutions of this system of equations are the next: φ1 = π φ3 = −φ2 ρ3 = ρ2 1. ρ1 = ρN2 2 − N2 + 2ρ2 cos φ2 = L ρ −ρ22 + 2 cos φ N 1 = K 2 2 ρ2
φ1 = φ2 = 0 φ3 = π ρ1 = ρ2Nρ3 2. ρ2 − ρ3 + ρ2Nρ3 = L ρ2 ρ3 + N ( ρ12 − ρ13 ) = −K
In an analogous way there might be found the integers L and K for which the first system has admissible solutions for λk . For N = 2 the same seven cases take place (ρ2 = ρ3 , φ1 = π and φ3 = −φ2 ): 1. L =r0, K = −2 p √ 3 1 −6= ρ2 = 1/3 3 46 + 6 57 + 12 √ √ 3 46+6
57
= 1.063200562, ρ1 = 1.7693, φ2 = 0.58798.
2. L =r1, K = −1 p √ 3 ρ2 = 1/3 3 44 + 3 177 + 21 √ 3
1 44+3
√
177
−3=
= 1.288008960, ρ1 = 1.20557, φ2 = 0.542940. 3. L =r0, K = −1 p √ 3 1 ρ2 = 1/3 3 53 + 6 78 + 3 √ −3= √ 3 53+6
78
= 1.146558421, ρ1 = 1.52138, φ2 = 0.845369 . 4. L = K = 0 ρ2 = 21/3 = 1.259921050, ρ1 = 1.259921050, φ2 = π/3. 5. L =r−1, K = 0 p √ 3 1 ρ2 = 1/3 3 54 + 6 87 − 18 √ = √ 3 54+6
87
= 1.086052036, ρ1 = 1.695620, φ2 = 1.2448. 6. L =r−1, K = 1 p √ 3 1 +3= ρ2 = 1/3 3 46 + 3 249 − 15 √ √ 3 46+3
249
= 1.215716761, ρ1 = 1.353211, φ2 = 1.425012. 7. L =r−2, K = 2 p √ 3 1 ρ2 = 1/3 3 26 + 6 33 − 24 √ +6= √ 3 26+6
33
= 1.138243270, ρ1 = 1.54369, φ2 = 1.77261. Following 3.2 we get the same condition for getting a diagonal matrix after three iterations: ρ1 = ρ2 = ρ3 = N 1/3 and 3φ1 = 3φ2 = 3φ3 ( mod 2π). This case corresponds to the integers L = K = 0. That is why the system for the integer elements of the matrix M will be the next: a+e+k =0 ak − gc − db + ek + ae − f h = 0 . aek − af h + dhc − dbk + gbf − gec = ±N 5. EXAMPLES OF M-D DECIMATION MATRICES
One of the widely used 2-D decimation matrix for twochannel systems is the quincunx matrix [3]: MQ =
·
1 1
1 −1
¸
,
where M2Q = 2I2 . The decimation matrix MQ is admis√ sible because its eigenvalues λ1 = −λ2 = i 2 > 1. The simplest possible 3-D decimation matrix for twochannel systems is the FCO - face centered orthorhombic matrix 1 0 1 MFCO = −1 −1 1 , 0 −1 0 √ where M3FCO = 2I3√ . The eigenvalues λ1 = 3 2, λ2 = √ 3 2exp(iπ/3), λ3 = 3 2 exp (−iπ/3).
For 4-D two-channel systems the sublattice is generated by [4] −1 −1 −1 −1 −1 0 −1 1 M4D = −1 1 1 −1 1 0 1 0 where M44D √ = 2I4 . The eigenvalues λ1 = √ √ 4 − 2, λ3 = i 4 2, λ4 = −i 4 2.
√ 4
2, λ2 =
6. CONCLUSION A method for generating all nonseparable admissible decimation matrices was developed for 2D and 3D cases. It allows to build any N -channel decimation matrix for which all requirements in 2 are met. The method might be generalized for any number of dimensions. 7. REFERENCES [1] A. Cohen and I. Daubechies, “Non-separable bidimensional wavelet bases,” Revista Matem. Iberoamericana, vol. 9, no. 1, pp. 51–137, 1993. [2] F. R. Gantmacher, Matrix theory, Nauka, Moscow, 550 p., 1988. [3] P. P. Vaidyanathan, Multirate Systems and Filter Banks, Prentice Hall, Englewood Cliffs, 1993. [4] O. Bolshakova, “Synthesis of four-dimensional filter banks by application of Bernstein polynomials,” in Proc. Fifth Intern. Conf. and Exhib. - Digit. Sign. Proc. and its Applic. - DSPA-2003, Moscow, 2003, MPEI, vol. 1, p. 278.
