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Wavelet Bases and Lifting Wavelets

Hongkai Xiong

熊红凯 http://min.sjtu.edu.cn

Department of Electronic Engineering

Shanghai Jiao Tong University

http://ivm.sjtu.edu.cn/

Wavelet Bases and Lifting Wavelets

• Wavelet Bases on an Interval

• Separable Wavelet Bases

• Lifting Wavelets

Wavelet Bases and Lifting Wavelets

• Wavelet Bases on an Interval

• Separable Wavelet Bases

• Lifting Wavelets

Wavelet Bases

Wavelets Over an Interval

Until now, only wavelets over the real axis have been considered, e.g.

wavelets that are suited to the analysis of signals defined over the whole axis.

In most cases, signals are compactly supported; images, in particular, are

explicitly defined over a rectangle of pixels.

The wavelets considered here are compactly supported.

A 0,𝑁 supported signal can be represented as the product of a general signal with the characteristic function of 0,𝑁 . The discontinuities of this function require special attention. Three methods are known to handle them:

wavelet periodization, wavelet folding and boundary wavelets, the last one

being the most efficient.

Wavelet Bases

Wavelet Periodization

A wavelet basis 𝜓𝑗,𝑛 𝑗,𝑛 ∈ℤ2 of 𝐋 𝟐 ℝ is transformed into a wavelet basis of 𝐋𝟐 0,1 by

periodizing each 𝜓𝑗,𝑛. The periodization of 𝑓 ∈ 𝐋 𝟐 ℝ over 0,1 is defined by

𝑓per 𝑡 = 𝑓 𝑡 + 𝑘

+∞

𝑘=−∞

The resulting periodic wavelets are

𝜓𝑗,𝑛 per

𝑡 = 1

2𝑗 𝜓

𝑡 − 2𝑗𝑛 + 𝑘

2𝑗

+∞

𝑘=−∞

For 𝑗 ≤ 0, there are 2−𝑗 different 𝜓𝑗,𝑛 per

indexed by 0 ≤ 𝑛 < 2−𝑗.

Wavelets whose support are completely inside the interval 0,1 are not changed:

𝜓𝑗,𝑛 per

𝑡 = 𝜓𝑗,𝑛 𝑡 for 𝑡 ∈ 0,1 .

The restriction to 0,1 of this periodization modifies only the boundary wavelets with a support that overlaps 𝑡 = 0 or 𝑡 = 1.

Wavelet Bases

Wavelet Periodization

As indicated in the Figure, wavelets that overlap the boundaries are

transformed into boundary wavelets that have two disjoint components

near 𝑡 = 0 and 𝑡 = 1. Taken separately, the components near 𝑡 = 0 and 𝑡 = 1 of these boundary wavelets have no vanishing moments, and thus create large signal coefficients.

Wavelet Bases

Wavelet Periodization

Theorem 7.16 For any 𝐽 ≤ 0,

𝜓𝑗,𝑛 per

−∞

Wavelet Bases

Wavelet Folding

To avoid creating discontinuous with wavelet periodization, the signal is folded

with respect to 𝑡 = 0: 𝑓0 𝑡 = 𝑓 𝑡 + 𝑓 −𝑡 . The support of 𝑓0 is −1,1 and it is transformed into a 2 periodic signal:

𝑓repl 𝑡 = 𝑓0 𝑡 − 2𝑘

+∞

𝑘=−∞

= 𝑓 𝑡 − 2𝑘

+∞

𝑘=−∞

+ 𝑓 2𝑘 − 𝑡

+∞

𝑘=−∞

This yields a continuous periodic signal:

The folded signal 𝑓repl 𝑡 is 2 periodic, symmetric about 𝑡 = 0 and 𝑡 = 1, and equal to 𝑓 𝑡 on 0,1

Wavelet Bases

Wavelet Folding

Decomposing 𝑓repl in a wavelet basis 𝜓𝑗,𝑛 𝑗,𝑛 ∈ℤ2 is equivalent to decomposing

𝑓 on a folded wavelet basis. Let 𝜓𝑗,𝑛 repl

be the folding of 𝜓𝑗,𝑛, one can verify that:

𝑓 𝑡 𝜓𝑗,𝑛

repl 𝑡

1

0

𝑑𝑡 = 𝑓repl +∞

−∞

𝑡 𝜓𝑗,𝑛 𝑡 𝑑𝑡

Suppose that 𝑓 is regular over 0,1 . Then 𝑓repl is continuous at 𝑡 = 0, 1 and produces smaller boundary wavelet coefficients than 𝑓per. However, it is not continuously differentiable at 𝑡 = 0, 1, which creates bigger wavelet coefficients at the boundary than inside.

The folded signal 𝑓repl 𝑡 is 2 periodic, symmetric about 𝑡 = 0 and 𝑡 = 1, and equal to 𝑓 𝑡 on 0,1

Wavelet Bases

Wavelet Folding

To construct a basis of 𝐋𝟐 0, 1 with the folded wavelets 𝜓𝑗,𝑛 repl

, it is sufficient

for 𝜓 𝑡 to be either symmetric or antisymmetric with respect to 𝑡 = 1/2.

The Haar wavelet is the only real compactly supported wavelet that is

symmetric or antisymmetric and that generates an orthogonal basis of 𝐋𝟐 ℝ .

If we loosen up the orthogonality constraint, then there exist biorthogonal

bases constructed with compactly supported wavelets that are either

symmetric or antisymmetric.

Wavelet Bases

Boundary Wavelets

Wavelet coefficients are small in regions where the signal is regular only if the

wavelets have enough vanishing moments.

The restriction of periodic and folded “boundary” wavelets to the

neighborhood of 𝑡 = 0 and 𝑡 = 1 have, respectively, 0 and 1 vanishing moments. Therefore, these “boundary” wavelets produce large inner products,

as if the signal were discontinuous or had a discontinuous derivative.

To avoid creating large-amplitude wavelet coefficients at the boundaries, one

must synthesize boundary wavelets that have as many vanishing moments as

the original wavelet 𝜓.

Wavelet Bases

Multiresolution of 𝐋𝟐 0, 1

A wavelet basis of 𝐋𝟐 0, 1 is constructed with a multiresolution

approximation 𝐕𝑗 int

−∞

Wavelet Bases

Multiresolution of 𝐋𝟐 0, 1

Consider an approximation space 𝐕𝑗 int ⊂ 𝐋𝟐 0, 1 with a compactly supported

Daubechies scaling function 𝜙 associated to a wavelet with 𝑝 vanishing moments. We translate 𝜙 so that its support is −𝑝 + 1, 𝑝 . At a scale 2𝑗 ≤ 2𝑝−1, there are 2−𝑗 − 2𝑝 scaling functions with a support inside 0, 1 :

𝜙𝑗,𝑛 int 𝑡 = 𝜙𝑗,𝑛 𝑡 =

1

2𝑗 𝜙

𝑡−2𝑗𝑛

2𝑗 for 𝑝 ≤ 𝑛 < 2−𝑗 − 𝑝

To construct an approximation space 𝐕𝑗 int of dimension 2−𝑗 we add 𝑝 scaling

functions with a support on the left boundary near 𝑡 = 0:

𝜙𝑗,𝑛 int 𝑡 =

1

2𝑗 𝜙𝑛 left 𝑡

2𝑗 for 0 ≤ 𝑛 < 𝑝

and 𝑝 scaling functions on the right boundary near 𝑡 = 1:

𝜙𝑗,𝑛 int 𝑡 =

1

2𝑗 𝜙 2−𝑗−1−𝑛

right 𝑡−1

2𝑗 for 2−𝑗 − 𝑝 ≤ 𝑛 < 2−𝑗

Wavelet Bases

Boundary effects are explicitly handled. Consider a Daubechies orthogonal basis with 𝑝

vanishing moments, the Strang and Fix condition 7.70 implies that there exists a

polynomial 𝑞𝑘 of degree 𝑘 such that:

𝑞𝑘 𝑡 = 𝑛

𝑘𝜙 𝑡 − 𝑛

+∞

𝑛=−∞

for 𝑘 < 𝑝

This equation is multiplied by the characteristic function of 0, 𝑁 . Assuming that the

support of 𝜙 is −𝑝 + 1, 𝑝 , scaling functions with indices 𝑝 ≤ 𝑘 < 𝑁 − 𝑝 are not

changed by this restriction. To recover the Strang and Fix condition on the interval, 𝑝 “left”

edge scaling function and 𝑝 “right ” edge scaling functions are to be found such that

𝑞𝑘 𝑡 𝟏 0,𝑁 𝑡 = 𝑛 𝑘𝜙 𝑡 − 𝑛 𝟏 0,𝑁 𝑡

+∞

𝑛=−∞

= 𝑎 𝑛

𝑝−1

𝑛=0

𝜙𝑛 left 𝑡 + 𝑛𝑘

𝑁−𝑝−1

𝑛=𝑝

𝜙 𝑡 − 𝑛 + 𝑏 𝑛

𝑝−1

𝑛=0

𝜙𝑛 right

𝑡

Boundary Wavelets

Wavelet Bases

𝑞𝑘 𝑡 𝟏 0,𝑁 𝑡 = 𝑛

𝑘𝜙 𝑡 − 𝑛 𝟏 0,𝑁 𝑡

+∞

𝑛=−∞

= 𝑎 𝑛

𝑝−1

𝑛=0

𝜙𝑛 left 𝑡 + 𝑛𝑘

𝑁−𝑝−1

𝑛=𝑝

𝜙 𝑡 − 𝑛 + 𝑏 𝑛

𝑝−1

𝑛=0

𝜙𝑛 right

𝑡

If this equation is satisfied, it remains valid after rescaling since the 𝑛𝑘, up to a

power of 2, are the scaling coefficients of 𝑞𝑘 at all resolutions. There remains to

find the filters ℎ and 𝐻 which satisfy the scaling equation:

𝜙𝑗,𝑘 int= 𝐻𝑘,𝑙

left𝜙𝑗+1,𝑙 int𝑝−1

𝑙=0 + ℎ𝑘,𝑚 left𝜙𝑗+1,𝑚

int𝑝+2𝑘 𝑚=𝑝

where 𝜙𝑗,𝑘 int 𝑡 denotes the whole set of scaling functions obtained by translation

at the resolution 𝑗, and to verify the orthogonality condition.

Boundary Wavelets

Wavelet Bases and Lifting Wavelets

• Wav