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Perfect Reconstruction FIR Filter Banks

Module by: Phil Schniter. E-mail the author

Summary: This module will drop the restrictive QMF conditions and focus on using FIR filters to achieve perfect reconstruction from filterbanks.

FIR Perfect-Reconstruction Conditions

The QMF design choices prevented the design of a useful (i.e., frequency selective) perfect-reconstruction (PR) FIR filterbank. This motivates us to re-examine PR filterbank design without the overly-restrictive QMF conditions. However, we will still require causal FIR filters with real coefficients.

Recalling that the two-channel filterbank (Figure 1),

Figure 1
Figure 1 (PR_f1.png)
has the input/output relation:
Y⁢z=12⁢( X⁢zX⁢-z )⁢( H 0 ⁢z H 1 ⁢z H 0 ⁢−z H 1 ⁢−z )⁢( G 0 ⁢z G 1 ⁢z ) Y z 1 2 X z X -z H 0 z H 1 z H 0 z H 1 z G 0 z G 1 z
(1)
we see that the delay-ll perfect reconstruction requires
( 2⁢z−l 0 )=( H 0 ⁢z H 1 ⁢z H 0 ⁢−z H 1 ⁢−z )⁢( G 0 ⁢z G 1 ⁢z ) 2 z l 0 H 0 z H 1 z H 0 z H 1 z G 0 z G 1 z
(2)
where H⁢z=( H 0 ⁢z H 1 ⁢z H 0 ⁢−z H 1 ⁢−z ) H z H 0 z H 1 z H 0 z H 1 z or, equivalently, that
( G 0 ⁢z G 1 ⁢z )=H -1z⁢( 2⁢z−l 0 )=1detH⁢z⁢( H 1 ⁢−z− H 1 ⁢z − H 0 ⁢−z H 0 ⁢z )⁢( 2⁢z−l 0 )=2detH⁢z⁢( z−l⁢ H 1 ⁢(−z) −(z−l⁢ H 0 ⁢−z) ) G 0 z G 1 z H z -1 2 z l 0 1 H z H 1 z H 1 z H 0 z H 0 z 2 z l 0 2 H z z l H 1 z z l H 0 z
(3)
where
detH⁢z= H 0 ⁢z⁢ H 1 ⁢−z− H 0 ⁢−z⁢ H 1 ⁢z H z H 0 z H 1 z H 0 z H 1 z
(4)
For FIR G 0 ⁢z G 0 z and G 1 ⁢z G 1 z , we require 1 that
detH⁢z=c⁢z−k H z c z k
(5)
for c∈R c and k∈Z k . Under this determinant condition, we find that
G 0 ⁢z G 1 ⁢z=2⁢z−(l−k)c⁢ H 1 ⁢−z− H 0 ⁢−z G 0 z G 1 z 2 z l k c H 1 z H 0 z
(6)
Assuming that H 0 ⁢z H 0 z and H 1 ⁢z H 1 z are causal with non-zero initial coefficient, we choose k=l k l to keep G 0 ⁢z G 0 z and G 1 ⁢z G 1 z causal and free of unnecessary delay.

Summary of Two-Channel FIR-PR Conditions

Summarizing the two-channel FIR-PR conditions: H 0 ⁢z& H 1 ⁢z=  causal real-coefficient  FIR H 0 z H 1 z   causal real-coefficient  FIR detH⁢z=c⁢z−l  ,   c∈R&l∈Z    c c l H z c z l G 0 ⁢z=2c⁢ H 1 ⁢−z G 0 z 2 c H 1 z G 1 ⁢z=-2c⁢ H 0 ⁢−z G 1 z -2 c H 0 z

Footnotes

  1. Since we cannot assume that FIR H 0 ⁢z H 0 z and H 1 ⁢z H 1 z share a common root.

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