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Perfect Reconstruction Filter Banks

We now consider filter banks with an arbitrary number of channels, and ask under what conditions do we obtain a perfect reconstruction filter bank? Polyphase analysis will give us the answer readily. Let's begin with the $ N$ -channel filter bank in Fig.11.20. The downsampling factor is $ R\leq N$ . For critical sampling, we set $ R=N$ .

Figure: $ N$ -channel filter bank.
\includegraphics[width=\twidth]{eps/FBNchan}

The next step is to expand each analysis filter $ H_k(z)$ into its $ N$ -channel ``type I'' polyphase representation:

$\displaystyle H_k(z) \eqsp \sum_{l=0}^{N-1} z^{-l} E_{kl}(z^N)$ (12.49)

or

$\displaystyle \underbrace{\left[\begin{array}{c} H_0(z) \\ [2pt] H_1(z) \\ [2pt] \vdots \\ [2pt] \!\!H_{N-1}(z)\!\!\end{array}\right]}_{\bold{h}(z)} = \underbrace{\left[\begin{array}{cccc} E_{0,0}(z^N) & E_{0,1}(z^N) & \cdots & E_{0,N-1}(z^N) \\ E_{1,0}(z^N) & E_{1,1}(z^N) & \cdots & E_{1,N-1}(z^N)\\ \vdots & \vdots & \cdots & \vdots\\ \!\!E_{N-1,0}(z^N) & E_{N-1,1}(z^N) & \cdots & E_{N-1,N-1}(z^N) \!\! \end{array}\right]}_{\bold{E}(z^N)} \underbrace{\left[\begin{array}{c} 1 \\ [2pt] z^{-1} \\ [2pt] \vdots \\ [2pt] \!\!z^{-(N-1)}\!\!\end{array}\right]}_{\bold{e}(z)}$ (12.50)

which we can write as

$\displaystyle \bold{h}(z) \eqsp \bold{E}(z^N)\bold{e}(z).$ (12.51)

Similarly, expand the synthesis filters in a type II polyphase decomposition:

$\displaystyle F_k(z) \eqsp \sum_{l=0}^{N-1} z^{-(N-l-1)}R_{lk}(z^N)$ (12.52)

or

$\displaystyle \underbrace{\left[\begin{array}{c} F_0(z) \\ [2pt] F_1(z) \\ [2pt] \vdots \\ [2pt] \!\!F_{N-1}(z)\!\!\end{array}\right]^T}_{\bold{f}^T(z)} \eqsp \underbrace{\left[\begin{array}{c} \!\!z^{-(N-1)}\!\! \\ [2pt] \!\!z^{-(N-2)}\!\! \\ [2pt] \vdots \\ [2pt] 1\end{array}\right]^T}_{{\tilde{\bold{e}}}(z)} \underbrace{\left[\begin{array}{cccc} R_{0,0}(z^N) & R_{0,1}(z^N) & \cdots & R_{0,N-1}(z^N) \\ R_{1,0}(z^N) & R_{1,1}(z^N) & \cdots & R_{1,N-1}(z^N)\\ \vdots & \vdots & \cdots & \vdots\\ \!\!R_{N-1,0}(z^N) & R_{N-1,1}(z^N) & \cdots & R_{N-1,N-1}(z^N)\!\! \end{array}\right]}_{\bold{R}(z^N)}$ (12.53)

which we can write as

$\displaystyle \bold{f}^T(z) \eqsp {\tilde{\bold{e}}}(z)\bold{R}(z^N).$ (12.54)

The polyphase representation can now be depicted as shown in Fig.11.21. When $ R=N$ , commuting the up/downsamplers gives the result shown in Fig.11.22. We call $ \bold{E}(z)$ the polyphase matrix.

Figure: Polyphase representation of the $ N$ -channel filter bank.
\includegraphics[width=\twidth]{eps/polyNchan}

Figure: Efficient polyphase form of the $ N$ -channel filter bank.
\includegraphics[width=\twidth]{eps/polyNchanfast}

As we will show below, the above simplification can be carried out more generally whenever $ R$ divides $ N$ (e.g., $ R=N/2, N/3,\ldots,
1$ ). In these cases $ \bold{E}(z)$ becomes $ \bold{E}(z^{N/R})$ and $ \bold{R}(z)$ becomes $ \bold{R}(z^{N/R})$ .



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``Spectral Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2011, ISBN 978-0-9745607-3-1.
Copyright © 2016-07-18 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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