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Spectral Audio Signal Processing
    Multirate Polyphase Filter Banks
       Perfect Reconstruction Filter Banks

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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.11.21. The downsampling factor is $ R\leq N$. For critical sampling, we set $ R=N$.

Figure: $ N$-channel filter bank.
\begin{figure}\input fig/FBNchan.pstex_t \end{figure}

The next step is to expand each analysis filter $ H_k(z)$ into its $ N$-channel ``Type 1'' polyphase representation:

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

or

$\displaystyle \underbrace{\left[\begin{array}{c} H_0(z) \\ [2pt] H_1(z) \\ [2pt... ...} \\ [2pt] \vdots \\ [2pt] \!\!z^{-(N-1)}\!\!\end{array}\right]}_{\bold{e}(z)} $

which we can write as

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

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

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

or

$\displaystyle \underbrace{\left[\begin{array}{c} F_0(z) \\ [2pt] F_1(z) \\ [2pt... ...-1,1}(z^N) & \cdots & R_{N-1,N-1}(z^N)\!\! \end{array}\right]}_{\bold{R}(z^N)} $

which we can write as

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

The polyphase representation can now be depicted as shown in Fig.11.22. When $ R=N$, commuting the up/downsamplers gives the result shown in Fig.11.23. We call $ \bold{E}(z)$ the polyphase matrix.

Figure: Polyphase representation of the $ N$-channel filter bank.
\begin{figure}\input fig/polyNchan.pstex_t \end{figure}

Figure: Efficient polyphase form of the $ N$-channel filter bank.
\begin{figure}\input fig/polyNchanfast.pstex_t \end{figure}

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})$.


Subsections

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Previous: Orthogonal Two-Channel Filter Banks
Next: Simple Examples of Perfect Reconstruction
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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