	Although DSP has long been considered an EE topic, recent developments have also generated signifi cant interest from the computer science community. DSP applications in the consumer market, such as bioinformatics, the MP3 audio format, and MPEG-based cable/satellite television have fueled a desire to understand this technology outside of hardware circles. Designed for upper division engineering and computer science students as well as practicing engineers, Digital Signal Processing Using MATLAB and Wavelets emphasizes the practical applications of signal processing. Over 100 MATLAB examples and wavelet techniques provide the latest applications of DSP, including image processing, games, fi lters, transforms, networking, parallel processing, and sound. The book also provides the mathematical processes and techniques needed to ensure an understanding of DSP theory. Designed to be incremental in diffi culty, the book will benefi t readers who are unfamiliar with complex mathematical topics or those limited in programming experience. Beginning with an introduction to MATLAB programming, it moves through filters, sinusoids, sampling, the Fourier transform, the z-transform and other key topics. An entire chapter is dedicated to the discussion of wavelets and their applications. A CD-ROM (platform independent) accompanies the book and contains source code, projects for each chapter, and the fi gures contained in the book. FEATURES: Contains over 100 short examples in MATLAB used throughout the book Includes an entire chapter on the wavelet transform Designed for the reader who does not have extensive math and programming experience Accompanied by a CD-ROM containing MATLAB examples, source code, projects, and fi gures from the book Contains modern applications of DSP and MATLAB project ideas

Digital Signal Processing Using MATLAB and Wavelets

By Michael Weeks

Chapter 9.3.2 - How The Haar Transform Affects a Point's Angle

To see how the angle changes for the Haar transform, we will start again with the point

, which we know becomes $\left(\frac{11}{\sqrt{2}}, \frac{-9}{\sqrt{2}}\right)$ in the Haar-domain. What are the angles that these points make with respect to their x-axes? The formula for finding the angles (labeled

) is given below. Note that atan finds the arctangent.

    if (x > 0) 
       theta = atan (y/x)  
    elseif (x < 0) 
       if (y < 0) 
          theta = atan (y/x) - pi
       else 
          theta = atan (y/x) + pi
       end 
    else 
       % The vector lies on the y-axis. 
       theta = sign(y)*(pi/2) 
    end

The angle made by the Cartesian coordinates

is $84.29^{\circ}$ . The angle made by the same coordinates after the Haar transform is $tan^{-1}\left(\frac{-9}{11}\right) = -39.29^{\circ}$ .

Looking at the relationship between the two in general, the $Angle_{Cartesian} = tan^{-1}\left(\frac{y}{x}\right)$ , while the $Angle_{Haar} = tan^{-1}\left(\frac{(x-y)/\sqrt{2}}{(x+y)/\sqrt{2}}\right)$ $= tan^{-1}\left(\frac{x-y}{x+y}\right)$ .

The formula above is rather cumbersome to deal with, due to the ambiguous nature of the arctan function, and because of a potential divide-by-zero problem when . For the first consideration, this becomes a little easier when we replace $x$ and $y$ with their polar coordinate equivalents. That is, we will replace and $y$ with $r \cos(\theta_1)$ and $r \sin(\theta_1)$ , respectively. The Haar-domain values are then $r (\cos(\theta_1) + \sin(\theta_1)) / \sqrt{2}$ and $r (\cos(\theta_1) - \sin(\theta_1)) / \sqrt{2}$ , respectively. Finding the new angle (the Haar-domain angle, $\theta_2$ ), we plug the Haar-domain values from above into the equation $\theta = tan^{-1}(\frac{y}{x})$ to find:

$\begin{displaymath}\theta_2 = tan^{-1}\left(\frac{r (\cos(\theta_1) - \sin(\thet... ...t{2}}{r( \cos(\theta_1) + \sin(\theta_1)) / \sqrt{2}}\right) . \end{displaymath}$

An interesting thing to notice is that the term $r/\sqrt{2}$ cancels itself out, meaning that the radii of the polar values do not matter. In other words, the value for $\theta_2$ depends solely on $\theta_1$ . This should be expected, since we already know from section 9.3.1 that the two radii are equal. So the equation for $\theta_2$ simplifies to:

$\begin{displaymath}\theta_2 = tan^{-1}\left(\frac{\cos(\theta_1) - \sin(\theta_1)}{\cos(\theta_1) + \sin(\theta_1)}\right) . \end{displaymath}$

We can divide by $\cos(\theta_1)$ , giving us:

$\begin{displaymath}\theta_2 = tan^{-1}\left(\frac{\frac{\cos(\theta_1)}{\cos(\th... ...cos(\theta_1)} + \frac{\sin(\theta_1)}{\cos(\theta_1)}}\right) \end{displaymath}$

$\begin{displaymath}\theta_2 = tan^{-1}\left(\frac{1 - \frac{\sin(\theta_1)}{\cos(\theta_1)}}{1 + \frac{\sin(\theta_1)}{\cos(\theta_1)}}\right) . \end{displaymath}$