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Spectral Audio Signal Processing
    Multirate Polyphase Filter Banks
       Critically Sampled Perfect Reconstruction Filter Banks
          Haar Example

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Haar Example

Before we leave this case (amplitude-complementary, two-channel, critically sampled, perfect reconstruction filter banks), let's see what happens when $ H_0(z)$ is the simplest possible lowpass filter having unity dc gain, i.e.,

$\displaystyle H_0(z) = \frac{1}{2} + \frac{1}{2}z^{-1}. $

This case is obtained above by setting $ E_0(z^2)=1/2$, $ o=1$, and $ h_0(1)=1/2$.

The polyphase components of $ H_0(z)$ are clearly

$\displaystyle E_0(z^2)=E_1(z^2)=1/2. $

Choosing $ H_1(z)=1-H_0(z)$ and choosing $ F_0(z)$ and $ F_1(z)$ for aliasing cancellation, the four filters become

\begin{eqnarray*} H_0(z) &=& \frac{1}{2} + \frac{1}{2}z^{-1} = E_0(z^2)+z^{-1}E_... ...F_1(z) &=& -H_0(-z) = -\frac{1}{2} + \frac{1}{2}z^{-1} = -H_1(z) \end{eqnarray*}

Thus, both the analysis and reconstruction filter banks are scalings of the familiar Haar filters (``sum and difference'' filters $ (1\pm z^{-1})/\sqrt{2}$).

The frequency responses are

\begin{eqnarray*} H_0(e^{j\omega}) &=& \quad\,F_0(e^{j\omega}) = \frac{1}{2} + \... ...ega}= j e^{-j\frac{\omega}{2}} \sin\left(\frac{\omega}{2}\right) \end{eqnarray*}

which are plotted in Fig.11.17.

Figure 11.17: Amplitude responses of the two channel filters in the Haar filter bank.
\includegraphics[width=3in]{eps/haar}

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Previous: Amplitude-Complementary 2-Channel Filter Bank
Next: Polyphase Decomposition of Haar Example
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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