# 4.2 The binary filter tree

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This module introduces the binary filter tree.

Recall that for image compression (see The 2-band Filter Bank ), the purpose of the 2-band filter bank in the Haar transform is tocompress most of the signal energy into the low-frequency band.

We may achieve greater compression if the low brand is further split into two. This may be repeated a number of times to givethe binary filter tree, shown with 4 levels in .

In 1-D, this is analogous to the way the 2-D Haar transform was extended to the multi-level Haar transform .

For an $N$ -sample input vector $x$ , the sizes and bandwidths of the signals of the 4-level filter tree are:

Signal No. of samples Approximate pass band
$x$ $N$ $0\to \frac{1}{2}{f}_{s}$
${y}_{1}$ $\frac{N}{2}$ $\frac{1}{4}\to \frac{1}{2}{f}_{s}$
${y}_{01}$ $\frac{N}{4}$ $\frac{1}{8}\to \frac{1}{4}{f}_{s}$
${y}_{001}$ $\frac{N}{8}$ $\frac{1}{16}\to \frac{1}{8}{f}_{s}$
${y}_{0001}$ $\frac{N}{16}$ $\frac{1}{32}\to \frac{1}{16}{f}_{s}$
${y}_{0000}$ $\frac{N}{16}$ $0\to \frac{1}{32}{f}_{s}$

Because of the downsampling (decimation) by 2 at each level, the total number of output samples = $N$ , regardless of the number of levels in the tree.

The ${H}_{0}$ filter is normally designed to be a lowpass filter with a passband from 0 to approximately $\frac{1}{4}$ of the input sampling frequency for that stage; and ${H}_{1}$ is a highpass (bandpass) filter with a pass bandapproximately from $\frac{1}{4}$ to $\frac{1}{2}$ of the input sampling frequency.

When formed into a 4-level tree, the filter outputs have the approximate pass bands given in . The final output ${y}_{0000}$ is a lowpass signal, while the other outputs are all bandpass signals, each covering a band of approximately oneoctave.

An inverse tree, mirroring , may be constructed using filters ${G}_{0}$ and ${G}_{1}$ instead of ${H}_{0}$ and ${H}_{1}$ , as shown for just one level in part (b) of this figure . If the PR conditions of this previous equation and this previous equation are satisfied, then the output of each level will be identical to the input of the equivalent level in , and the final output will be a perfect reconstruction of the input signal.

## Multi-rate filtering theorem

To calculate the impulse and frequency responses for a multistage network with downsampling at each stage, as in , we must first derive an important theorem for multirate filters.

The downsample-filter-upsample operation of (a) is equivalent to either the filter-downsample-upsample operation of (b) or the downsample-upsample-filter operation of (c), if the filter is changed from $H(z)$ to $H(z^{2})$ .

From (a):

Take z-transforms:
Reverse the order of summation and let $m=n-2i$ : therefore,
where $Y(z)=H(z^{2})X(z)$

This describes the operations of (b). Hence the first result is proved.

The result from line 3 in

$\stackrel{^}{Y}(z)=\frac{1}{2}(X(z)+X(-z))H(z^{2})=\stackrel{^}{X}(z)H(z^{2})$
shows that the filter $H(z^{2})$ may be placed after the down/up-sampler as in (c), which proves the second result.

## General results for m:1 subsampling

It can be shown that:

• $H(z)$ becomes $H(z^{M})$ if shifted ahead of an M:1 downsampler or following an M:1 upsampler.
• M:1 down/up-sampling of a signal $X(z)$ produces:
$\stackrel{^}{X}(z)=\frac{1}{M}\sum_{m=0}^{M-1} X(ze^{\frac{i\times 2\pi m}{M}})$

## Transformation of the filter tree

Using the result of , can be redrawn as in with all downsamplers moved to the outputs. (Note requires much more computation than .)

We can now calculate the transfer function to each output (before the downsamplers) as:

${H}_{01}(z)={H}_{0}(z){H}_{1}(z^{2})$
${H}_{001}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{1}(z^{4})$
${H}_{0001}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{0}(z^{4}){H}_{1}(z^{8})$
${H}_{0000}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{0}(z^{4}){H}_{0}(z^{8})$
In general the transfer functions to the two outputs at level $k$ of the tree are given by:
${H}_{k,1}=\prod_{i=0}^{k-2} {H}_{0}(z^{2^{i}}){H}_{1}(z^{2^{(k-1)}})$
${H}_{k,0}=\prod_{i=0}^{k-1} {H}_{0}(z^{2^{i}})$
For the Haar filters of this equation and this equation from our discussion of the 2-band filter bank, the transfer functionsto the outputs of the 4-level tree become:
${H}_{01}(z)=\frac{1}{2}(z^{-3}+z^{-2}-z^{-1}+1)$
${H}_{001}(z)=\frac{1}{2\sqrt{2}}(z^{-7}+z^{-6}+z^{-5}+z^{-4}-z^{-3}+z^{-2}+z^{-1}+1)$
${H}_{0001}(z)=\frac{1}{4}(z^{-15}+z^{-14}+z^{-13}+z^{-12}+z^{-11}+z^{-10}+z^{-9}+z^{-8}-z^{-7}+z^{-6}+z^{-5}+z^{-4}+z^{-3}+z^{-2}+z^{-1}+1)$
${H}_{0000}(z)=\frac{1}{4}(z^{-15}+z^{-14}+z^{-13}+z^{-12}+z^{-11}+z^{-10}+z^{-9}+z^{-8}+z^{-7}+z^{-6}+z^{-5}+z^{-4}+z^{-3}+z^{-2}+z^{-1}+1)$

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