The International Archives of the Photogrammetry, Remote Sensing and Spatial Information Sciences. Vol. XXXVII. Part B7. Beijing 2008 
FREQUENCY RESPONSES 
Figure 2. The frequency responses of \ and h^ in the 
example 
To show this, suppose that\(«),fy(«), and h 2 (n) satisfy the 
PR conditions and that 
h 2 {n) = \{N-l-n). (8) 
Then, by defining 
V e » = 
V e » = 
^(^(») + h 2 (n-2d)), 
-j=(h^(n) - ^(n - 2d)) with de Z, 
(9) 
(10) 
the filters , hf ew , and h 2 new also satisfy the PR conditions, 
and h ] new , h 2 new are symmetric and anti-symmetric as follows: 
V-to-Vitfa-l-»*), 
h 2 new ( n ) = -h 2 new (N 2 -l-n), 
where N 2 = N + 2d . 
We state main results of the paper (Selesnick, 2004) without the 
proof. The filters h 0 , \ , and h 2 with symmetries Eq. (7) and 
Eq. (8) satisfy the PR conditions if their polyphase components 
of the filters are given by 
H 0 ,o( z ) = z~ M ' 2 j2A(z)B(z~ ] ), 
(ID 
H l0 (z) = A 2 (z), 
(12) 
H\ ti (z) = -B 2 (z), 
(13) 
2) Find A(z) and B(z) from H 00 (z) and U(z) by 
using factorization and root selection. 
3) Find // 10 (z)and /F,,(z)by using Eq. (12) and Eq. 
(13), respectively. 
4) Find Hfz) and H 2 (z) by using // 10 (z) , H n (z) 
and Eq. (8). 
5) Obtain (anti-) symmetric wavelets h x and h 2 by using 
Eq. (9) and Eq. (10). 
Figure 3. The symmetric scaling function, <j>{t), and the two 
wavelets y/\t) and ^ 2 (/)of the example 
n 
^o(«) 
\(n) 
h 2 {n) 
0 
0.00069616789827 
-0.00014203017443 
0.00014203017443 
1 
-0.02692519074183 
0.00549320005590 
-0.00549320005590 
2 
-0.04145457368920 
0.01098019299363 
-0.00927404236573 
3 
0.19056483888763 
-0.13644909765612 
0.07046152309968 
4 
0.58422553883167 
-0.21696226276259 
0.13542356651691 
5 
0.58422553883167 
0.33707999754362 
-0.64578354990472 
6 
0.19056483888763 
0.33707999754362 
0.64578354990472 
7 
-0.04145457368920 
-0.21696226276259 
-0.13542356651691 
8 
-0.02692519074183 
-0.13644909765612 
-0.07046152309968 
9 
0.00069616789827 
0.01098019299363 
0.00927404236573 
10 
0 
0.00549320005590 
0.00549320005590 
1 
0 
-0.00014203017443 
-0.00014203017443 
Table 1. Coefficients for the example 
2.3.4 Example: To obtain a lowpass filter, h 0 (n) , with a 
minimal length, we used a maximally flat lowpass even-length 
FIR filter with the following transfer function : 
1 + z 
-i\ 
z + 2 + z 
m + k - 0.5 V -z + 2 - z -1 
k 
where 
A(z)A(z~') = 0.5 + 0.5 U(z), 
B(z)B(z~') = 0.5 - 0.5U(z), 
C/(z)i/(z _l ) = l-2i/ 00 (z)// 00 (z -1 ), and M = N/2 - 1. 
2.3.3 Filter Design: First, obtain a lowpass filter, h 0 {n) , 
which has an even-length and which satisfies the symmetry 
condition of Eq. (6). The design procedure is as follows: 
1) Knowing H 0 (z) and, thus H 00 (z) , use spectral 
factorization to find U(z) from 1 - 2H 0 0 (z)H 0 0 (z"‘). 
Unfortunately, although the setting of H 0 (z) := F m n (z) gives an 
Hfz) that does not satisfy Eq. (6), we can use a linear 
combination of various F m n (z) values to obtain a H 0 (z) filter 
that does satisfy Eq. (6). For example, if we use a setting of 
H 0 {z) = z^faF^z) + (1 - a)F 3>1 (z)), (14) 
then, for special values of a , H 0 (z) satisfies Eq. (6).