4.7 Bi-orthogonal perfect reconstruction fir filterbanks

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.
• $z^{-(N-1)}F(z)=H_0(z)H_1(-z)$

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: $$G_i=\{a_i, \overline{a_i}, \frac{1}{a_i}, \frac{1}{\overline{a_i}}\}()$$ $$\forall a_i, \left|a_i\right|=1\colon G_i=\{a_i, \overline{a_i}\}()$$ $$\forall a_i, a_i\in \mathbb{R}\colon G_i=\{a_i, \frac{1}{a_i}\}()$$ $$\forall a_i, a_i=\pm (1)\colon G_i=\{a_i\}()$$ Choose
  Note that ${\widehat{H}}_0(z)$ and ${\widehat{H}}_1(z)$ 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 ${\widehat{H}}_0(z)$ from those roots. Then construct ${\widehat{H}}_1(-z)$ from the roots in the remaining root groups. Finally,construct ${\widehat{H}}_1(z)$ from ${\widehat{H}}_1(-z)$ by reversing the signs of odd-indexed coefficients.
• ${\widehat{H}}_0(z)$ and ${\widehat{H}}_1(z)$ are the desired analysis filters up to a scaling. To take care of the scaling, first create ${\tilde{H}}_0(z)=a{\widehat{H}}_0(z)$ and ${\tilde{H}}_1(z)=b{\widehat{H}}_1(z)$ where $a$ and $b$ are selected so that $\sum {\tilde{h}}_0(n)=1=\sum {\tilde{h}}_1(n)$ . Then create $H_0(z)=c{\tilde{H}}_0(z)$ and $H_1(z)=c{\tilde{H}}_1(z)$ 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 $\sum h_0(n)\sum h_1(n)-1^m=1$ .