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 below:

• The downsampling factor is $R\leq N$ .
• For critical sampling, $R=N$ .

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

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

or

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

which we can write as

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

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

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

or

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

which we can write as

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

The polyphase representation can now be depicted as

When $R=N$ , commuting the up/downsamplers gives

We call $\bold{E}(z)$ the polyphase matrix.

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