Create NxN Haar Matrix

Question

I can't find a definition for generating an un-normalized NxN Haar matrix. So what is the equation?

see: http://en.wikipedia.org/wiki/Haar_wavelet

Thx, Chris

The Wikipedia article gives the definition. What part is giving you trouble? — Teepeemm, May 26 '14 at 13:38

score 4 · Answer 1 · edited Nov 02 '17 at 04:07

Here is the algorithm for both the normalized and un-normalized Haar matrix based on the recursive formula for the Haar matrix:

from the paper "Discrete wavelets and perturbation theory" by W-H Steeb, et al. Here is the implementation in Python

def haarMatrix(n, normalized=False):
    # Allow only size n of power 2
    n = 2**np.ceil(np.log2(n))
    if n > 2:
        h = haarMatrix(n / 2)
    else:
        return np.array([[1, 1], [1, -1]])

    # calculate upper haar part
    h_n = np.kron(h, [1, 1])
    # calculate lower haar part 
    if normalized:
        h_i = np.sqrt(n/2)*np.kron(np.eye(len(h)), [1, -1])
    else:
        h_i = np.kron(np.eye(len(h)), [1, -1])
    # combine parts
    h = np.vstack((h_n, h_i))
    return h

score 1 · Answer 2 · answered May 26 '14 at 14:55

It depends on what exactly you want to achieve. The Haar matrix is the 2x2 DCT matrix, so inversly, you can treat the NxN DCT(II) matrix as the Haar matrix for that block size.

Or if the N is dyadic, N=2^n, then you might be asking for the transform matrix for n stages of the Haar transform. Which might be a problem because of the sampling rate decimation in each step.

score 1 · Accepted Answer · edited Feb 09 '17 at 15:09

1

Thx all. Wikipedia gives the 'equation':

$H_{2N} = \begin{bmatrix} H_{N} \otimes [1, 1] \\ I_{N} \otimes [1, -1] \end{bmatrix} $$ :where $$ I_{N} = \begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix}$

I've wrote a recursive solution for generating an un-normalized NxN Haar matrix in octave.

function [h] = haar(n)
h = [1];
if n > 2
    h = haar(n/2);
endif
% calculate upper haar part
h_n = kron(h,[1,1]); 
% calculate lower haar part 
h_i = kron(eye(length(h)),[1,-1]);
% combine parts
h = [h_n; h_i];
endfunction

disp(haar(8));

edited Feb 09 '17 at 15:09

Community

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answered May 27 '14 at 16:53

user2372976

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score 0 · Answer 4 · answered Dec 23 '21 at 12:43

def HT(N):
    if N == 1: return np.array([[1]])
    return 1/np.sqrt(2)*np.concatenate(
        (
            np.kron(HT(N//2),[1,1])
        ,
            np.kron(np.identity(N//2),[1,-1])
        ),axis = 0
        )

I used the equation in Wikipedia adding a normalization factor 1/sqrt(2). The matrix becomes orthonormal. Does it work?

Create NxN Haar Matrix

4 Answers4