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Wavelet Dilation Equation

Module by: Ivan Selesnick. E-mail the author

A Condition for Existence

Dilation Equation

φt=2nhnφ2tn φ t 2 n h n φ 2 t n
(1)
φtdt=2nhnφ2tndt t φ t t 2 n h n φ 2 t n
(2)
φtdt=2nhnφ2tndt t φ t 2 n h n t φ 2 t n
(3)
φtdt=2nhnφ2tdt t φ t 2 n h n t φ 2 t
(4)
φtdt=2nhnφ2tdt t φ t 2 n h n t φ 2 t
(5)
φtdt=2nhn12φtdt t φ t 2 n h n 1 2 t φ t
(6)
1=12nhn 1 1 2 n h n
(7)
assuming φtdt0 t φ t 0 . For a solution φt φ t to even exist, it is required that the coefficients hn h n add up to 2 2 .
nhn=2 n h n 2
(8)

Fourier Form of the Dilation Equation

Dilation Equation

φt=2nhnφ2tn φ t 2 n h n φ 2 t n
(9)
LHS=RHS LHS RHS φt=2nhnφ2tn φ t 2 n h n φ 2 t n Φω=2nhnφ2tn Φ ω 2 n h n φ 2 t n Φω=2nhnφ2tne(iωt)dt Φ ω 2 n h n t φ 2 t n ω t For τ=2tn τ 2 t n , Φω=2nhn12φτe(iωτ+n2)dτ Φ ω 2 n h n 1 2 τ φ τ ω τ n 2 Φω=12nhne(iω2n)φτe(iω2τ)dτ Φ ω 1 2 n h n ω 2 n τ φ τ ω 2 τ Φω=12nhne(iω2n)Φω2 Φ ω 1 2 n h n ω 2 n Φ ω 2 Φω=12Φω2nhne(iω2n) Φ ω 1 2 Φ ω 2 n h n ω 2 n Note that H f ω=DTFThn=nhne(iωn) H f ω DTFT h n n h n ω n is the discrete-time Fourier transform (DTFT) of the scaling filter hn h n . We then have the Fourier form of the dilation equation:
Φω=12 H f ω2Φω2 Φ ω 1 2 H f ω 2 Φ ω 2
(10)

When ω=0 ω 0 is put into this equation, we get Φ0=12 H f 0Φ0 Φ 0 1 2 H f 0 Φ 0 or

H f 0=2 H f 0 2
(11)
which is exactly the same equation as nhn=2 n h n 2 that we already got.

From Equation 10, we can write Φω2=12 H f ω4Φω4 Φ ω 2 1 2 H f ω 4 Φ ω 4 or Φω=12 H f ω212 H f ω4Φω4 Φ ω 1 2 H f ω 2 1 2 H f ω 4 Φ ω 4 If we iterate, we get the infinite-product formula for Φω Φ ω ,

Φω=Φ0k=112 H f ω2k Φ ω Φ 0 k 1 1 2 H f ω 2 k
(12)
From this formula, we can see that the zeros of Φω Φ ω are determined by the zeros of H f ω H f ω .

What is the support of φ(t)?

Given hn h n , what is the support of φt φ t ?

Suppose hn h n is supported on 0nN1 0 n N 1 then from the dilation equation φt=2n=0N1hnφ2tn φ t 2 n 0 N 1 h n φ 2 t n the support of φt φ t is found by matching the support of the left hand side and the right hand side.

Suppose the support of φt φ t is a b a b . Then

  • φ2t φ 2 t has support a2 b2 a 2 b 2
  • φ2t1 φ 2 t 1 has support a+12 b+12 a 1 2 b 1 2
  • φ2tk φ 2 t k has support a+k2 b+k2 a k 2 b k 2
  • φ2t(N1) φ 2 t N 1 has support a+N12 b+N12 a N 1 2 b N 1 2
The support of the LHS is a b a b by assumption. The support of the RHS is a2 b2+N12 a 2 b 2 N 1 2 . Therefore, matching the endpoints we get a=a2 a a 2 b=b2+N12 b b 2 N 1 2 which gives a=0 a 0 b=N1 b N 1 The support of φt φ t is 0 N1 0 N 1 . hn  has finite support  φt  of finite support   h n   has finite support   φ t   of finite support   That means that there are finite degrees of freedom in design of finitely supported scaling functions.

Sum-Integral Equality of φ(t)

Note: limit  j 12jlφl2j=φtdt j 1 2 j l φ l 2 j t φ t Let us define j,j1: S j 12jlφl2j j j 1 S j 1 2 j l φ l 2 j Begin with the dilation equation φt=2nhnφ2tn φ t 2 n h n φ 2 t n and substitute l2j l 2 j for tt. φl2j=2nhnφl2j1n φ l 2 j 2 n h n φ l 2 j 1 n with j1 j 1 . Now sum over ll: lφl2j=l2nhnφl2j1n l φ l 2 j l 2 n h n φ l 2 j 1 n lφl2j=2nhnlφl2j1n l φ l 2 j 2 n h n l φ l 2 j 1 n lφl2j=2nhnlφl2j1 l φ l 2 j 2 n h n l φ l 2 j 1 lφl2j=2lφl2j1 l φ l 2 j 2 l φ l 2 j 1 Divide both sides by 2j 2 j : 12jlφl2j=12j1lφl2j1 1 2 j l φ l 2 j 1 2 j 1 l φ l 2 j 1 S j = S j 1 S j S j 1 For j1 j 1 , all the S j S j are the same! limit  j S j = S 0 j S j S 0 Therefore

φtdt=kφk t φ t k φ k
(13)

If φt φ t is continuous, then it can also be shown that

φtdt=kφ t o +k t φ t k φ t o k
(14)
for any real t o t o .

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