Numerical recipes

13.10 Wavelet Transforms 593
p.” This results in the output of H, decimated by half, accurately representing the
data’s “smooth” information. The output of G, also decimated, is referred to as
the data’s “detail” information [4].
For such a characterization to be useful, it must be possible to reconstruct the
original data vector of length N from its N/2 smooth or s-components and its N/2
detail or d-components. That is effected by requiring the matrix (13.10.1) to be
orthogonal, so that its inverse is just the transposed matrix
⎡
c0 c3 ··· c2 c1
⎤
⎢ c1
⎢ c2
⎢ c3
⎢
⎣
−c2
c1
−c0
c0
c1
c3
−c2
···
. ..
c2
c3
c1
−c0
c0
c1
c3
−c2
c3 −c0 ⎥
⎦
c2 c1 c0 c3
c3 −c0 c1 −c2
(13.10.2)
One sees immediately that matrix (13.10.2) is inverse to matrix (13.10.1) if and
only if these two equations hold,
c 2 0 + c2 1 + c2 2 + c2 3 =1
c2c0 + c3c1 =0
(13.10.3)
If additionally we require the approximation condition of order p =2, then two
additional relations are required,
c3 − c2 + c1 − c0 =0
(13.10.4)
0c3 − 1c2 +2c1 − 3c0 =0
Equations (13.10.3) and (13.10.4) are 4 equations for the 4 unknowns c 0,...,c3,
first recognized and solved by Daubechies. The unique solution (up to a left-right
reversal) is
c0 =(1+ √ 3)/4 √ 2 c1 =(3+ √ 3)/4 √ 2
c2 =(3− √ 3)/4 √ 2 c3 =(1− √ 3)/4 √ 2
(13.10.5)
In fact, DAUB4 is only the most compact of a sequence of wavelet sets: If we had
six coefficients instead of four, there would be three orthogonality requirements in
equation (13.10.3) (with offsets of zero, two, and four), and we could require the
vanishing of p =3moments in equation (13.10.4). In this case, DAUB6, the solution
coefficients can also be expressed in closed form,
c0 =(1+ √ 10 + � 5+2 √ 10)/16 √ 2 c1 =(5+ √ 10 + 3 � 5+2 √ 10)/16 √ 2
c2 =(10−2 √ �
10 + 2 5+2 √ 10)/16 √ 2 c3 =(10−2 √ �
10 − 2 5+2 √ 10)/16 √ 2
c4 =(5+ √ 10 − 3 � 5+2 √ 10)/16 √ 2 c5 =(1+ √ 10 − � 5+2 √ 10)/16 √ 2
(13.10.6)
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