Editing Haar wavelet (section)

==Haar matrix==
The 2×2 Haar matrix that is associated with the Haar wavelet is
: <math> H_2 = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.</math>
Using the [[discrete wavelet transform]], one can transform any sequence <math>(a_0,a_1,\dots,a_{2n},a_{2n+1})</math> of even length into a sequence of two-component-vectors <math> \left(\left(a_0,a_1\right),\left(a_2,a_3\right),\dots,\left(a_{2n},a_{2n+1}\right)\right) </math>. If one right-multiplies each vector with the matrix <math> H_2 </math>, one gets the result <math>\left(\left(s_0,d_0\right),\dots,\left(s_n,d_n\right)\right)</math> of one stage of the fast Haar-wavelet transform. Usually one separates the sequences ''s'' and ''d'' and continues with transforming the sequence ''s''. Sequence ''s'' is often referred to as the ''averages'' part, whereas ''d'' is known as the ''details'' part.<ref>{{cite book |first1=David K. |last1=Ruch |first2=Patrick J. |last2=Van Fleet |title=Wavelet Theory: An Elementary Approach with Applications |year=2009 |publisher=John Wiley & Sons|isbn=978-0-470-38840-2 }}</ref>

If one has a sequence of length a multiple of four, one can build blocks of 4 elements and transform them in a similar manner with the 4×4 Haar matrix
: <math> H_4 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 0 & 0\\ 0 & 0 & 1 & -1 \end{bmatrix},</math>
which combines two stages of the fast Haar-wavelet transform.

Compare with a [[Walsh matrix]], which is a non-localized 1/–1 matrix.

Generally, the 2N×2N Haar matrix can be derived by the following equation.

: <math> H_{2N} = \begin{bmatrix} H_{N} \otimes [1, 1] \\ I_{N} \otimes [1, -1] \end{bmatrix}</math>
:where <math>I_{N} = \begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix}</math> and <math>\otimes</math> is the [[Kronecker product]].

The [[Kronecker product]] of <math>A \otimes B</math>, where <math>A</math> is an m×n matrix and <math>B</math> is a p×q matrix, is expressed as

: <math>A \otimes B = \begin{bmatrix} a_{11}B & \dots & a_{1n}B \\ \vdots & \ddots & \vdots \\ a_{m1}B & \dots & a_{mn}B\end{bmatrix}.</math>

An un-normalized 8-point Haar matrix <math>H_8</math> is shown below

: <math>H_{8} = \begin{bmatrix} 1&1&1&1&1&1&1&1 \\ 1&1&1&1&-1&-1&-1&-1 \\ 1&1&-1&-1&0&0&0&0& \\ 0&0&0&0&1&1&-1&-1 \\ 1&-1&0&0&0&0&0&0& \\ 0&0&1&-1&0&0&0&0 \\ 0&0&0&0&1&-1&0&0& \\ 0&0&0&0&0&0&1&-1 \end{bmatrix}.</math>

Note that, the above matrix is an un-normalized Haar matrix. The Haar matrix required by the Haar transform should be normalized.

From the definition of the Haar matrix <math>H</math>, one can observe that, unlike the [[Fourier transform]], <math>H</math> has only real elements (i.e., 1, -1 or 0) and is non-symmetric.

Take the 8-point Haar matrix <math>H_8</math> as an example. The first row of <math>H_8</math> measures the average value, and the second row of <math>H_8</math> measures a low frequency component of the input vector. The next two rows are sensitive to the first and second half of the input vector respectively, which corresponds to moderate frequency components. The remaining four rows are sensitive to the four section of the input vector, which corresponds to high frequency components.<ref>{{cite web |url=http://fourier.eng.hmc.edu/e161/lectures/Haar/index.html |title=haar |publisher=Fourier.eng.hmc.edu |date=30 October 2013 |access-date=23 November 2013 |archive-date=21 August 2012 |archive-url=https://web.archive.org/web/20120821004423/http://fourier.eng.hmc.edu/e161/lectures/Haar/index.html |url-status=dead }}</ref>