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Course by: Phil Schniter. E-mail the author

# Orthogonal Perfect Reconstruction FIR Filterbank

Module by: Phil Schniter. E-mail the author

Summary: This module introduces the ideas behind and design issues of Orthogonal Perfect Reconstruction of FIR filterbanks.

## Orthogonal PR Filterbanks

The FIR perfect-reconstruction (PR) conditions leave some freedom in the choice of H 0 z H 0 z and H 1 z H 1 z . Orthogonal PR filterbanks are defined by causal real-coefficient even-length-NN analysis filters that satisfy the following two equations:

1= H 0 z H 0 z-1+ H 0 -z H 0 z-1 1 H 0 z H 0 z H 0 -z H 0 z
(1)
H 1 z=±z(N1) H 0 z-1 H 1 z ± z N 1 H 0 z
(2)
To verify that these design choices satisfy the FIR-PR requirements for H 0 z H 0 z and H 1 z H 1 z , we evaluate det H z H z under the second condition above. This yields
det H z=± H 0 z H 1 z-1 H 0 -z H 1 z=(z(N1))( H 0 z H 0 z-1+ H 0 -z H 0 z-1)=z(N1) H z ± H 0 z H 1 z H 0 -z H 1 z z N 1 H 0 z H 0 z H 0 -z H 0 z z N 1
(3)
which corresponds to c=-1 c -1 and l=N1 l N 1 in the FIR-PR determinant condition det H z=czl H z c z l . The remaining FIR-PR conditions then imply that the synthesis filters are given by
G 0 z=-2 H 1 z-1=2z(N1) H 0 z-1 G 0 z -2 H 1 z 2 z N 1 H 0 z
(4)
G 1 z=2 H 0 z=2z(N1) H 1 z-1 G 1 z 2 H 0 z 2 z N 1 H 1 z
(5)
The orthogonal PR design rules imply that H 0 ejω H 0 ω is "power symmetric" and that H 0 ejω H 1 ejω H 0 ω H 1 ω form a "power complementary" pair. To see the power symmetry, we rewrite the first design rule using z=ejω z ω and -1=e±jπ -1 ± π , which gives
1= H 0 ejω H 0 e(jω)+ H 0 ej(ωπ) H 0 e(j)(ωπ)=| H 0 ejω|2+| H 0 ej(ωπ)|2=| H 0 ejω|2+| H 0 ej(πω)|2 1 H 0 ω H 0 ω H 0 ω H 0 ω H 0 ω 2 H 0 ω 2 H 0 ω 2 H 0 ω 2
(6)
The last two steps leveraged the fact that the DTFT of a real-coefficient filter is conjugate-symmetric. The power-symmetry property is illustrated in Figure 1:

Power complementarity follows from the second orthogonal PR design rule, which implies | H 1 ejω|=| H 0 ej(πω)| H 1 ω H 0 ω . Plugging this into the previous equation, we find

1=| H 0 ejω|2+| H 1 ejω|2 1 H 0 ω 2 H 1 ω 2
(7)
The power-complimentary property is illustrated in Figure 2:

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