# 4.7 Bi-orthogonal perfect reconstruction fir filterbanks

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This module looks at Bi-Orthogonal PR-FIR filterbanks and shows how they are similar to orthogonal designs yet provide linear-phase filters.

## Bi-orthogonal filter banks

Due to the minimum-phase spectral factorization, orthogonal PR-FIR filterbanks will not have linear-phase analysis and synthesis filters. Non-linear phase may be undesirable forcertain applications. "Bi-orthogonal" designs are closely related to orthogonal designs, yet give linear-phase filters.The analysis-filter design rules for the bi-orthogonal case are

• $F(z)$ : zero-phase real-coefficient halfband such that $F(z)=\sum_{n=-(N-1)}^{N-1} f(n)z^{-n}$ , where $N$ is even.
• $z^{-(N-1)}F(z)={H}_{0}(z){H}_{1}(-z)$
It is straightforward to verify that these design choices satisfy the FIR perfect reconstruction condition $\det H(z)=cz^{-l}$ with $c=1$ and $l=N-1$ :
$\det H(z)={H}_{0}(z){H}_{1}(-z)-{H}_{0}(-z){H}_{1}(z)=z^{-(N-1)}F(z)--1^{-(N-1)}z^{-(N-1)}F(-z)=z^{-(N-1)}(F(z)+F(-z))=z^{-(N-1)}$
Furthermore, note that $z^{-(N-1)}F(z)$ is causal with real coefficients, so that both ${H}_{0}(z)$ and ${H}_{1}(z)$ can be made causal with real coefficients. (This was another PR-FIR requirement.) The choice $c=1$ implies that the synthesis filters should obey ${G}_{0}(z)=2{H}_{1}(-z)$ ${G}_{1}(z)=-2{H}_{0}(-z)$ From the design choices above, we can see that bi-orthogonal analysis filter design reduces to the factorization of acausal halfband filter $z^{-(N-1)}F(z)$ into ${H}_{0}(z)$ and ${H}_{1}(z)$ that have both real coefficients and linear-phase. Earlier we saw thatlinear-phase corresponds to root symmetry across the unit circle in the complex plane, and that real-coefficientscorrespond to complex-conjugate root symmetry. Simultaneous satisfaction of these two properties can be accomplished by quadruples of roots. However, there are special cases in which a root pair, or even a single root, cansimultaneously satisfy these properties. Examples are illustrated in :

The design procedure for the analysis filters of a bi-orthogonal perfect-reconstruction FIR filterbank issummarized below:

• Design a zero-phase real-coefficient filter $F(z)=\sum_{n=-(N-1)}^{N-1} f(n)z^{-n}$ where N is a positive even integer (via, e.g. , window designs, LS, or equiripple).
• Compute the roots of $F(z)$ and partition into a set of root groups $\{{G}_{0}, {G}_{1}, {G}_{2}, \dots \}()$ that have both complex-conjugate and unit-circle symmetries. Thus a root group may have one ofthe following forms: ${G}_{i}=\{{a}_{i}, \overline{{a}_{i}}, \frac{1}{{a}_{i}}, \frac{1}{\overline{{a}_{i}}}\}()$ $\forall {a}_{i}, \left|{a}_{i}\right|=1\colon {G}_{i}=\{{a}_{i}, \overline{{a}_{i}}\}()$ $\forall {a}_{i}, {a}_{i}\in \mathbb{R}\colon {G}_{i}=\{{a}_{i}, \frac{1}{{a}_{i}}\}()$ $\forall {a}_{i}, {a}_{i}=±(1)\colon {G}_{i}=\{{a}_{i}\}()$ Choose
Note that ${\stackrel{^}{H}}_{0}(z)$ and ${\stackrel{^}{H}}_{1}(z)$ will be real-coefficient linear-phase regardless of which groups are allocated to whichfilter. Their frequency selectivity, however, will be strongly influenced by group allocation. Thus, you manyneed to experiment with different allocations to find the best highpass/lowpass combination. Note also thatthe length of ${H}_{0}(z)$ may differ from the length of ${H}_{0}(z)$ .
a subset of root groups and construct ${\stackrel{^}{H}}_{0}(z)$ from those roots. Then construct ${\stackrel{^}{H}}_{1}(-z)$ from the roots in the remaining root groups. Finally,construct ${\stackrel{^}{H}}_{1}(z)$ from ${\stackrel{^}{H}}_{1}(-z)$ by reversing the signs of odd-indexed coefficients.
• ${\stackrel{^}{H}}_{0}(z)$ and ${\stackrel{^}{H}}_{1}(z)$ are the desired analysis filters up to a scaling. To take care of the scaling, first create ${\stackrel{~}{H}}_{0}(z)=a{\stackrel{^}{H}}_{0}(z)$ and ${\stackrel{~}{H}}_{1}(z)=b{\stackrel{^}{H}}_{1}(z)$ where $a$ and $b$ are selected so that $\sum {\stackrel{~}{h}}_{0}(n)=1=\sum {\stackrel{~}{h}}_{1}(n)$ . Then create ${H}_{0}(z)=c{\stackrel{~}{H}}_{0}(z)$ and ${H}_{1}(z)=c{\stackrel{~}{H}}_{1}(z)$ where $c$ is selected so that the property $z^{-(N-1)}F(z)={H}_{0}(z){H}_{1}(-z)$ is satisfied at DC ( i.e. , $z=e^{i\times 0}=1$ ). In other words, find $c$ so that $\sum {h}_{0}(n)\sum {h}_{1}(n)-1^{m}=1$ .

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