	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 9.5.3 - Down-Sampling and Up-Sampling with Daubechies 4

This is why the coefficients from the Daubechies four-coefficient wavelet, when used without the down-sampler and up-sampler, ended up with $y[n] = 2 x[n-3]$ . If the down-samplers and up-samplers are included, then the 2 term drops out. This is shown later. First, we will start, as usual, with Figure 9.12, an updated filter bank for four coefficients in each filter.

**Figure 9.12:** A filter bank with four taps per filter.

Signals $w[n]$ and $z[n]$ are just as they were earlier:

$\begin{displaymath}w[n] = ax[n] + bx[n-1] + cx[n-2] + dx[n-3] \end{displaymath}$

$\begin{displaymath}z[n] = dx[n] - cx[n-1] + bx[n-2] - ax[n-3] . \end{displaymath}$

Signals $w[n-k]$ and $z[n-k]$ , are again found by replacing $n$ with $n-k$ :

$\begin{displaymath}w[n-k] = ax[n-k] + bx[n-k-1] + cx[n-k-2] + dx[n-k-3] \end{displaymath}$

$\begin{displaymath}z[n-k] = dx[n-k] - cx[n-k-1] + bx[n-k-2] - ax[n-k-3] . \end{displaymath}$

Looking at $w_d[n]$ and $z_d[n]$ , we again have to be careful about whether $n$ is even or odd. When $n$ is odd, there is no value for or !

$\begin{displaymath}w_d[n] = w[n], \;\; n\;\mathrm{is\;even} \end{displaymath}$

$\begin{displaymath}z_d[n] = z[n], \;\; n\;\mathrm{is\;even} \end{displaymath}$

Signals $w_u[n]$ and $z_u[n]$ must be treated with equal caution:

$\begin{displaymath}w_u[n] = w_d[n] = w[n], \;\; n\;\mathrm{is\;even} \end{displaymath}$

$\begin{displaymath}w_u[n] = 0, \;\; n\;\mathrm{is\;odd} \end{displaymath}$

$\begin{displaymath}z_u[n] = z_d[n] = z[n], \;\; n\;\mathrm{is\;even} \end{displaymath}$

$\begin{displaymath}z_u[n] = 0, \;\; n\; \mathrm{is \; odd.} \end{displaymath}$

We have values for $w_u[n]$ and $z_u[n]$ for even and odd values of n. So when using these signals for $n$ and $n-1$ , one of indices must be even, while the other is odd. Likewise, if $n$ is even, then so is $n-2$ , $n-4$ , etc. The final signals of each channel are:

$\begin{displaymath}w_f[n] = d w_u[n] + c w_u[n-1] + b w_u[n-2] + a w_u[n-3] \end{displaymath}$

$\begin{displaymath}w_f[n] = d w[n] + 0 + b w[n-2] + 0, \;\; n\; \mathrm{is \; even} \end{displaymath}$

$\begin{displaymath}w_f[n] = 0 + c w[n-1] + 0 + a w[n-3], \;\; n\; \mathrm{is \; odd} \end{displaymath}$

$\begin{displaymath}z_f[n] = -a z_u[n] + b z_u[n-1] - c z_u[n-2] + d z_u[n-3] \end{displaymath}$

$\begin{displaymath}z_f[n] = -a z[n] + 0 - c z[n-2] + 0, \;\; n\; \mathrm{is \; even} \end{displaymath}$

$\begin{displaymath}z_f[n] = 0 + b z[n-1] + 0 + d z[n-3], \;\; n\; \mathrm{is \; odd.} \end{displaymath}$

Keeping track of whether index $n$ is even or odd, we can find $y[n] = w_f[n] + z_f[n]$ :

$\begin{displaymath}y[n] = d w[n] + b w[n-2] - a z[n] - c z[n-2], \;\; n\; \mathrm{is \; even} \end{displaymath}$

$\begin{displaymath}y[n] = c w[n-1] + a w[n-3] + b z[n-1] + d z[n-3], \;\; n\; \mathrm{is \; odd.} \end{displaymath}$

These expressions can be expanded, to put the output $y[n]$ in terms of the original input $x[n]$ :

$\begin{eqnarray*} y[n] &=& d(a x[n] + b x[n-1] + c x[n-2] + d x[n-3]) \& & ... ...- c x[n-4] + b x[n-5] - a x[n-6]), \;\; n\; \mathrm{is \; odd.} \end{eqnarray*}$

Simplifying:

$\begin{eqnarray*} y[n] &=& ad x[n] + bd x[n-1] + cd x[n-2] + dd x[n-3] \& &... ...cd x[n-4] + bd x[n-5] - ad x[n-6], \;\; n\; \mathrm{is \; odd.} \end{eqnarray*}$

Eliminating terms that cancel each other out, and simplifying further:

$\begin{eqnarray*} y[n] &=& ac x[n-1] + bd x[n-1] \& & + aa x[n-3] + bb x[n-... ... \& & + ac x[n-5] + bd x[n-5], \;\; n\; \mathrm{is \; odd.} \end{eqnarray*}$

In this final step, we notice two things. First, the expressions are the same regardless of whether $n$ is even or odd, so now we can represent it as a single statement:

$\begin{eqnarray*} y[n] &=& aa x[n-3] + bb x[n-3] + cc x[n-3] + dd x[n-3] \& & + ac x[n-1] + bd x[n-1] + ac x[n-5] + bd x[n-5] . \end{eqnarray*}$