Polyphase Decomposition of Haar Example

Let's look at the polyphase representation for this example. Starting with the filter bank and its reconstruction (see Fig.10.18), the polyphase decomposition of $H_0(z)$ is

$$H_0(z) = E_0(z^2) + z^{-1}E_1(z^2) = \frac{1}{2}+\frac{1}{2}z^{-1}$$

Thus, $E_0(z^2)=E_1(z^2)=1/2$ , and therefore

$$H_1(z) = 1-H_0(z) = E_0(z^2)-z^{-1}E_1(z^2).$$

We may derive polyphase synthesis filters as follows:

$$\begin{eqnarray*} \hat{X}(z) &=& \left[F_0(z)H_0(z) + F_1(z)H_1(z)\right] X(z)\\... ...)-H_1(z)\right] + z^{-1}\left[H_0(z) + H_1(z)\right]\right\}X(z) \end{eqnarray*}$$

The polyphase representation of the filter bank and its reconstruction can now be drawn as in Fig.10.19. Notice that the reconstruction filter bank is formally the transpose of the analysis filter bank [231]. A filter bank that is inverted by its own transpose is said to be an orthogonal filter bank, a subject to which we will return §10.3.8.

Figure 10.19: Polyphase representation of the general two-channel, critically sampled filter bank and its inverse.

Figure 10.20: Figure 10.19 with downsamplers commuted inside the filter branches.

Commuting the downsamplers (using the noble identities from §10.2.5), we obtain Figure 10.20. Since $E_0(z)=E_1(z)=1/2$ , this is simply the OLA form of an STFT filter bank for $N=2$ , with $N=M=R=2$ , and rectangular window $w=[1/2,1/2]$ . That is, the DFT size, window length, and hop size are all 2, and both the DFT and its inverse are simply sum-and-difference operations.