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The connection between orthogonal wavelets and orthogonal filters through the concept of multiresolution analysis. It covers the orthonormality of scaling functions, the refinement equation, and the double shift orthogonality condition. The document also discusses the importance of orthogonal filters for the existence of orthogonal wavelets.
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2
2D Vector Space:
Basis vectors are i, j ���� orthonormal basis
y
x
v
x
v
y
v
v
x
and v
y
are the projections of v onto the x and
y axes:
v
x
= ���� v , i ���� i
v
y
� v , j � ��
� j
� ��
� ��
� ��
���
� v , i � = [v
x
v
y
= v
x
Inner Product
Orthogonal multiresolution spaces:
j
has an orthonormal basis {
j/
j
t œ k): - � � k � �}
j,k
(t)
j
has an orthonormal basis {
j/
w(
j
t œ k): -� � k � �}
w
j,k
(t)
3
Orthonormal means
j,k
(t) , �
j,l
(t)� =
�
j/
j
t œ k) 2
j/
j
t - l)dt = �[k - l]
�w
j,k
(t) , w
j,l
(t)� =
�
j/
w(
j
t œ k) 2
j/
w(
j
t - l)dt = �[k - l]
For orthogonal multiresolution spaces, we have
� �� �
j
j
So
j,k
(t) , w
j,l
(t)� = 0
4
��� ��� ��� ��� ��� ���
� ��
� ��
� ��
� ��
� ��
� ��
� ��
Biorthogonal Wavelet Bases
Two scaling functions and two wavelets:
Synthesis:
�(t) = 2 �f
0
[k] �(2t œ k)
k
w(t) = 2 �f
1
[k] �(2t œ k)
k
Analysis:
�(t) = 2 � h
0
[-k] �(2t œ k)
k
w(t) = 2 � h
1
[-k] �(2t œ k)
k
7
Two sets of multiresolution spaces:
{0} � … V
0
� V
1
� …� V
j
� … � L
2
(�)
{0} � … V
0
� V
1
� … � V
j
� … � L
2
(�)
V
j
j
= V
j+
V
j
j
= V
j+
Spaces are orthogonal w.r.t. each other i.e.
V
j
h W
j
h
� k � ��
��
�}
V
j
W
j
V
0
has a basis {�(t œ k) :
V
0
has a basis {�(t œ k) : - � � k � �}
W
0
has a basis {w(t œ k): - � � k � �}
W
0
has a basis {w(t œ k): - � � k � �}
8
���
���
�
Bases are orthogonal w.r.t. each other i.e.
�� (t) �
(t œ k) dt = �[k] �� (t) w(t œ k)dt = 0
�w(t) � (t œ k) dt = 0 �w(t) w(t œ k)dt = �[k]
Equivalent to perfect reconstruction conditions on filters
Representation of functions in a biorthogonal basis:
f(t) = � c
k
�(t œ k) + � � d
j,k
j/
w(
j
t œ k)
k j=0 k
c
k
= � f(t) �(t œ k) dt
d
j,k
j/
� f(t) w(
j
t œ k) dt
9
Similarly, we can represent f(t) in the dual basis
�
~
f(t) = � c
k
�(t œ k) + � � d
j,k
j/
w(
j
t œ k)
k j=0 k
c
k
= � f(t) �(t œ k) dt
d
j,k
j/
� f(t) w(
j
t œ k) dt
Note: When f
0
[k] = h
0
[-k] and f
1
[k] = h
1
[-k], we have
�(t) = �
(t) � V
j
j
w(t) = w(t) � W
j
j
i.e. we have orthogonal wavelets!
10
���
� ��
���
���
���
���
��� ���
���
���
���
���
���
���
���
���
���
���
�
�
�
�
�
�
�
Note: with the alternating flip requirement, which was
necessary for the highpass filter in the case of
orthogonal filters, we can show that
� w(t) w(t œ n)dt = �[n]
and
� �(t) w(t œ n)dt = 0
13
Orthogonality in the Frequency Domain
Let
a[n] = � �(t) �(t œ n)dt
Use Parseval‘s theorem
a[n] =
1
2 ����
����
(�) e
-i�n
d� = ��� ������
2
����
1
2 � ��
�
e
i�n
d�
Trick: split integral over entire � axis into a sum of
integrals over 2� intervals
a[n] =
1
2 ����
(� + 2�k)�
2
e
i(� + 2�k)n
d�
k=-�
�� �� ���� ���� ���� (���� + 2����k)����
2
1
2 ����
= e
i�n
d�
k=-�
14
���
���
���
� ��
� ��
���
���
�
�
Take the Discrete Time Fourier Transform of both sides
A(�) � � a[n]e
œi�n
�
(� + 2�k)�
2
n k=-�
If �(t œ n) are orthonormal, then a[n] = �[n]
So the scaling function and it shifts are orthogonal if
���� ��������(���� + 2����k)����
2
Note: if we set ����
�� ��
k=-� ��
�
= 0, then
(2�k)�
2
k
and since �
(0) = 1, we see that
(2�k) = 0 for k � 0
15
Connection between orthogonal wavelets and
orthogonal filters (in frequency domain):
Start with an orthogonal scaling function:
1 = � ��(� + 2�k)�
2
k=-�
and then change scale using the refinement equation in
the frequency domain:
0
0
2
�� ( �/2 + �k)�
2
k=-�
16
