Digital Signal Processing Using MATLAB and Wavelets

Chapter 9.3 - How the Haar Transform Is a 45-Degree Rotation

A basis is a projection onto a coordinate axis. If we have a point on a graph, we can talk of its distance from the origin, another point used for reference. But this only narrows it down to a circle, so we need more information: the angle between the vector made by the origin to the point of interest, and another line for reference (i.e., the x-axis). For a point in 2D space, these polar coordinates are sufficient information to unambiguously locate the point with respect to the origin. But we often use the Cartesian basis {1, 0} and {0, 1} to represent this point, which tells about the point in terms of a unit vector along the x-axis and another unit vector along the y-axis (they are normalized). We know that the y-axis intersects the x-axis at the origin, and that there is a 90-degree angle between them (they are orthogonal). Though, generally speaking, we could use a different set of axes (i.e., a different basis).

The Haar basis is { $1/\sqrt{2}$, $1/\sqrt{2}$ } and {$1/\sqrt{2}$, $-1/\sqrt{2}$ }, instead of the normal Cartesian basis. This section answers the question of what the points look like in the Haar-domain instead of the Cartesian-domain. What we will see is that the points have the same relationship to one another, but the coordinate system moves by a rotation of 45 degrees.

First, consider the points $(x_1, y_1), (x_2, y_2), etc.$ as the array $\{ x_1, y_1, x_2, y_2, etc. \}$. The Cartesian system simply isolates the $x$'s and $y$'s, as in:


\begin{displaymath}\left[
\begin{array}{ccc}
\left[ 1 \; 0 \right] & 0 \; 0 \\...
...gin{array}{c}
x_1 \\ y_1 \\ x_2 \\ y_2
\end{array} \right] . \end{displaymath}

Whereas the Haar system does the following:


\begin{displaymath}\frac{1}{\sqrt{2}} \left[
\begin{array}{cccc}
\left[1 \; 1\...
...gin{array}{c}
x_1 \\ y_1 \\ x_2 \\ y_2
\end{array} \right] . \end{displaymath}

For example, the Cartesian coordinates $(1, 10)$ would be mapped to $\left(\frac{1 + 10}{\sqrt{2}}, \frac{1 - 10}{\sqrt{2}}\right)$, or $\left(\frac{11}{\sqrt{2}}, \frac{-9}{\sqrt{2}}\right)$ in the Haar-domain.

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