Wavelet Bases and Lifting Wavelets - wavelet and ¢  Periodic wavelet bases have the disadvantage

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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