Algorithm of Wavelet Transform (with Qt class)

Leave a comment

There are many kind of wavelet transform, and I think the names are quite confusing.

For instance, there are continuous and discrete wavelet transforms, in which, the “continuous” and “discrete” are for the wavelet parameters, not for the “data” itself. Therefore, for discrete data, there are “continuous” and “discrete” wavelet transforms, and for function, there are also “continuous” and “discrete” wavelet transforms.

In here, we will focus on discrete wavelet transform for function first. This discrete wavelet transform is also called as wavelet series, which express a compact support function into series of wavelet.

For simplicity, we also focus on orthonormal wavelet.

As the wavelet span the entire space, any compact function can be expressed as

\displaystyle f(t) = \sum_{j,k} \left<f(t)|\psi_{j,k}(t)\right> \psi_{j,k}(t)

\psi_{j,k}(t) = 2^{j/2} \psi(2^j t - k)

where j, k are integer.


Now, we move to discrete data discrete wavelet transform. The data is discrete, we can imagine only t_n = t_0 + n \Delta points are known with finite n .

\displaystyle f_n = f(t_n) = \sum_{j,k} \left<f_n|\psi_{j,k}(t_n) \right> \psi_{j,k}(t_n)

the integration becomes a finite sum.

Without loss of generality, we can set t_0 = 0, \Delta = 1, and then the time axis becomes an integer number axis. We found that j  \leq 0 as the wavelet can only be expand, not shrink. Because there are finite number of data point, i.e. n < \infty, -Log_2(n) < j \leq 0 .

However, this double summation for each f_n is very time consuming. There is a Fast Discrete Wavelet Transform. Before we continuous, we must study the wavelet.


From the last post, we know that the scaling function that generate a MRA must be:

\displaystyle \phi(t) = \sum_{k} g_0(k) \phi(2t-k)

\left<\phi(t-k) | \phi(t-k') \right> = \delta_{kk'}

, where k are integer. The set of shifted scaling function span a space V_0 . For the wavelet,

\displaystyle \psi(t) = \sum_{k} g_1(k) \psi(2t-k)

\left<\psi(t-k) | \psi(t-k') \right> = \delta_{kk'}

The set of shifted wavelet span a space W_0, so that W_0 \perp V_0, so that

\left<\phi(t-k)|\psi(t-k') \right> = 0

Since the wavelet is generated from the scaling function, we expect the coefficient of g_0(k) and g_1(k) are related. In fact, the relationship for orthonormal scaling function and wavelet is

g_1(k) = (-1)^k g_0(1-k)


For discrete data x_i , it can be decomposed into the MRA space. We start by the largest V_0 space, where the wavelet is most shrunken.

\displaystyle x_i = \sum_{k} v_{0,k} \phi(i-k)

to decompose to the V_{-1} and W_{-1} space. We can use the nested property of the MRA space, \phi(2t) can be decomposed into \phi(t-k) and \psi(t-k) ,

\displaystyle \psi(2t-l) = \sum_{k} h_0(2k-l) \phi(t-k) + h_1(2k-l) \psi(t-k)

where (given that \phi(t) and $\latex \psi(t)$ are orthonormal ),

h_0(2k-l) = \left< \phi(2t-l) | \phi(t-k) \right>

h_1(2k-l) = \left< \phi(2t-l) | \psi(t-k) \right>

Therefore, using the coefficient of h_0 and h_1, the wavelet coefficient v_{0,k} can be decomposed to

\displaystyle v_{s-1,k} = \sum_{l} h_0(2k-l) v_{s,l}  

\displaystyle w_{s-1,k} = \sum_{l} h_1(2k-l) v_{s,l}

in graphic representation

h1.PNG

This is a fast discrete wavelet transform.


Due to the nested space of MRA, we also expect that the coefficient h_0 and h_1 are related to g_0 . For orthonormal wavelet,

\displaystyle h_0(k) = \frac{1}{2} g_0(-k)

\displaystyle h_1(k) = \frac{1}{2} (-1)^{k} g_0 (k+1)

Since the g_0 is finite, the g_1, h_0, h_1 are all finite. That greatly reduce the computation cost of the discrete wavelet transform.


To reconstruct the discrete data x_i, we don’t need to use

\displaystyle v_{s+1,l} = \sum_{k} v_{s,k} \phi(l - k) + w_{s,k} \psi(l-k)

using the nested space of MRA, \psi(t) = \sum_{k} g_1(k) \psi(2t-k) ,

\displaystyle v_{s+1,l} = \sum_{k} g_0(l-2k) v_{s,k} + g_1(l-2k) w_{s,k}

in graphical representation,

h2.PNG


I attached the wavelet transfrom class for Qt, feel free to modify.

https://drive.google.com/file/d/0BycN9tiDv9kmMVRfcVFMbDFWS0k/view?usp=sharing

https://drive.google.com/file/d/0BycN9tiDv9kmUmlmNk1kaVJCbEU/view?usp=sharing

in the code, the data did not transform to MRA space. The code treats the data already in the MRA space. Some people said this is a “crime”. But for the seek of “speed”, it is no need to map the original discrete data into MRA space. But i agree, for continuous function, we must map to MRA space.

 

 

Changing of frame II

Leave a comment

Few things have to say in advance.

  1. A Vector is NOT its coordinate
  2. A vector can only be coordinated when there is a frame.
  3. A frame is a set of “reference” vectors, which span the whole space. Those reference vectors are called basis of a frame.
  4. a transformation is on a vector or its coordinate. And it can be represented by a matrix.
  5. A Matrix should act on a coordinate or basis, but not a vector.

where

\hat{\alpha} = \begin {pmatrix} \hat{\alpha_1} \\ . \\ \hat{\alpha_n} \end{pmatrix} is the column vector of  basis reference vector.

\vec{u_{\alpha}} is the coordinate column vector in \alpha basis.

\vec{U} is the vector in space

\vec{V} is the transformed vector in space.

G and H are the matrix of transform.

G \cdot H \cdot G^{-1} has the same meaning of H , only the matrix representation of the transform is different due to different basis.

the Euler’s rotation can be illustrated by series of the diagram. each rotation of frame can be made by each G . but when doing real calculation, after we apply the matrix G  on the coordinate, the basis changed. when we using the fact that  a matrix can be regard as a frame transform or vector transform. we have follow:

This diagram can extend to any series of frame rotation. and the V_s \rightarrow X_s \rightarrow V_2 \rightarrow V_s triangle just demonstrate how 2 steps frame transform can be reduced to the vector transform in same frame.

i finally feel that i understand Euler angle and changing of frame fully. :D

HERE is a note on vector transform and frame transform.