Digital Signal Processing Using MATLAB and Wavelets

Chapter 9.5.2 - Down-Sampling and Up-Sampling with 2 Coefficients

With wavelets, the filter bank structure will include down-samplers and up-samplers in the following configuration, Figure 9.11. In each channel, the signal undergoes several changes. To keep them distinct, they are each labeled. In the upper channel, we have $z[n]$, which is the output from the first or "analyzing' FIR filter, then $z_d[n]$ corresponds to $z[n]$ after it has been down-sampled, then $z_u[n]$ represents the signal after up-sampling, and finally $z_f[n]$ gives us the final values for that channel.

Figure 9.11: A two-channel filter bank with down/up-samplers.


Signals $w[n]$ and $z[n]$ are just as they were in section 9.1:

\begin{displaymath}w[n] = ax[n] + bx[n-1] \end{displaymath}

\begin{displaymath}z[n] = bx[n] - ax[n-1] . \end{displaymath}

Signals $w[n-1]$ and $z[n-1]$, are again found by replacing $n$ above with $n-1$:

\begin{displaymath}w[n-1] = a x[n-1] + b x[n-2] \end{displaymath}

\begin{displaymath}z[n-1] = b x[n-1] - a x[n-2] . \end{displaymath}

Looking at $w_d[n]$ and $z_d[n]$, we have to be a bit careful. We make the statements below for the even values of $n$. When $n$ is odd, there is no value for $w_d[n]$ or $z_d[n]$!

\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 the indices must be even, while the other is odd. The final signals of each channel are:

\begin{displaymath}w_f[n] = b w_u[n] + a w_u[n-1] \end{displaymath}

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

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

\begin{displaymath}z_f[n] = -a z_u[n] + b z_u[n-1] \end{displaymath}

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

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

When we combine $w_f[n]$ and $z_f[n]$ to form $y[n]$, we still have to concern ourselves with whether the index is even or odd.

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

\begin{displaymath}y[n] = a w[n-1] + b z[n-1], \;\; 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{displaymath}y[n] = b (a x[n] + b x[n-1]) - a (b x[n] - a x[n-1]), \;\; n\;\mathrm{is\;even} \end{displaymath}

\begin{displaymath}y[n] = a (a x[n-1] + b x[n-2]) + b (b x[n-1] - a x[n-2]), \;\; n\;\mathrm{is\;odd} \end{displaymath}

Simplifying:

\begin{displaymath}y[n] = ab x[n] + bb x[n-1] - ab x[n] + aa x[n-1], \;\; n\;\mathrm{is\;even} \end{displaymath}

\begin{displaymath}y[n] = aa x[n-1] + ab x[n-2] + bb x[n-1] - ab x[n-2], \;\; n\;\mathrm{is\;odd.} \end{displaymath}

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

\begin{displaymath}y[n] = (aa + bb) x[n-1], \;\; n\;\mathrm{is\;even} \end{displaymath}

\begin{displaymath}y[n] = (aa + bb) x[n-1], \;\; n\;\mathrm{is\;odd.} \end{displaymath}

This last step is important, where we see that $y[n]$ is in terms of $x[n-1]$ regardless of whether $n$ is even or odd. If we compare this result to the result of section 9.1, also dealing with Haar coefficients, we see that the difference is a factor of 2. The Haar coefficients are usually given as $1/\sqrt{2}$, $1/\sqrt{2}$, so that $(aa + bb) = 1$.

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