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Two-Channel Critically Sampled Filter Banks

Figure 11.15 shows a simple two-channel band-splitting filter bank, followed by the corresponding synthesis filter bank which reconstructs the original signal (we hope) from the two channels. The analysis filter $ H_0(z)$ is a half-band lowpass filter, and $ H_1(z)$ is a complementary half-band highpass filter. The synthesis filters $ F_0(z)$ and $ F_1(z)$ are to be derived. Intuitively, we expect $ F_0(z)$ to be a lowpass that rejects the upper half-band due to the upsampler by 2, and $ F_1(z)$ should do the same but then also reposition its output band as the upper half-band, which can be accomplished by selecting the upper of the two spectral images in the upsampler output.

% latex2html id marker 30377\psfrag{x(n)}{\normalsize $x(n)$}\psfrag{x}{\normalsize $x$}\psfrag{(n)}{\normalsize $(n)$}\psfrag{y}{\normalsize $y$}\psfrag{v}{\normalsize $v$}\psfrag{H}{\normalsize $H$}\psfrag{F}{\normalsize $F$}\psfrag{z}{\normalsize $z$}\begin{figure}[htbp]
\caption{Two-channel critically sampled filter bank.}

The outputs of the two analysis filters in Fig.11.15 are

$\displaystyle X_k(z) \eqsp H_k(z)X(z), \quad k=0,1.$ (12.16)

Using the results of §11.1, the signals become, after downsampling,

$\displaystyle V_k(z) \eqsp \frac{1}{2}\left[X_k(z^{1/2}) + X_k(-z^{1/2})\right], \; k=0,1.$ (12.17)

After upsampling, the signals become
$\displaystyle Y_k(z) \eqsp V_k(z^2)$ $\displaystyle =$ $\displaystyle \frac{1}{2}[X_k(z) + X_k(-z)]$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}[H_k(z)X(z) + H_k(-z)X(-z)],\; k=0,1.$  

After substitutions and rearranging, we find that the output $ \hat{x}$ is a filtered replica of the input signal plus an aliasing term:
$\displaystyle \hat{X}(z)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[H_0(z)F_0(z) + H_1(z)F_1(z)\right]X(z)$  
  $\displaystyle +$ $\displaystyle \frac{1}{2}\left[H_0(-z)F_0(z) + H_1(-z)F_1(z)\right]X(-z)
\protect$ (12.18)

For perfect reconstruction, we require the aliasing term to be zero. For ideal half-band filters cutting off at $ \omega=\pi/2$ , we can choose $ F_0=H_0$ and $ F_1=H_1$ and the aliasing term is zero because there is no spectral overlap between the channels, i.e., $ H_0(-z)F_0(z)=H_1(z)H_0(z)=0$ , and $ H_1(-z)F_1(z)=H_0(z)H_1(z)=0$ . However, more generally (and more practically), we can force the aliasing to zero by choosing synthesis filters
$\displaystyle F_0(z)$ $\displaystyle =$ $\displaystyle \quad\! H_1(-z),$  
$\displaystyle F_1(z)$ $\displaystyle =$ $\displaystyle -H_0(-z).
\protect$ (12.19)

In this case, synthesis filter $ F_0(z)$ is still a lowpass, but the particular one obtained by $ \pi$ -rotating the highpass analysis filter around the unit circle in the $ z$ plane. Similarly, synthesis filter $ F_1(z)$ is the $ \pi$ -rotation (and negation) of the analysis lowpass filter $ H_0(z)$ on the unit circle. For this choice of synthesis filters $ F_0$ and $ F_1$ , aliasing is completely canceled for any choice of analysis filters $ H_0$ and $ H_1$ .

Referring again to (11.18), we see that we also need the non-aliased term to be of the form

$\displaystyle A(z)$ $\displaystyle =$ $\displaystyle H_0(z)F_0(z) + H_1(z)F_1(z)
\protect$ (12.20)

where $ A(z)$ is of the form

$\displaystyle A(z)\eqsp g\,z^{-d}.$ (12.21)

That is, for perfect reconstruction, we need, in addition to aliasing cancellation, that the non-aliasing term reduce to a constant gain $ g$ and/or delay $ d$ . We will call this the filtering cancellation constraint on the channel filters. Thus perfect reconstruction requires both aliasing cancellation and filtering cancellation.

Let $ {\tilde H}$ denote $ H(-z)$ . Then both constraints can be expressed in matrix form as follows:

$\displaystyle \left[\begin{array}{cc} H_0 & H_1 \\ [2pt] {\tilde H}_0 & {\tilde H}_1 \end{array}\right]\left[\begin{array}{c} F_0 \\ [2pt] F_1 \end{array}\right]\eqsp \left[\begin{array}{c} c \\ [2pt] 0 \end{array}\right]$ (12.22)

Substituting the aliasing-canceling choices for $ F_0$ and $ F_1$ from (11.19) into the filtering-cancellation constraint (11.20), we obtain

$\displaystyle g\,z^{-d}$ $\displaystyle =$ $\displaystyle H_0(z)H_1(-z) - H_1(z)H_0(-z).
\protect$ (12.23)

The filtering-cancellation constraint is almost satisfied by ideal zero-phase half-band filters cutting off at $ \pi/2$ , since in that case we have $ H_1(-z)=H_0(z)$ and $ H_0(-z)=H_1(z)$ . However, the minus sign in (11.23) means there is a discontinuous sign flip as frequency crosses $ \omega=\pi/2$ , which is not equivalent to a linear phase term. Therefore the filtering cancellation constraint fails for the ideal half-band filter bank! Recall from above, however, that ideal half-band filters did work using a different choice of synthesis filters, relying instead on their lack of spectral overlap. The presently studied case from (11.19) arose from so-called Quadrature Mirror Filters (QMF), which are discussed further below. First, however, we'll look at some simple special cases.

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``Spectral Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2011, ISBN 978-0-9745607-3-1.
Copyright © 2022-02-28 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University