The Biquad Allpass Section

The general biquad transfer function was given in Eq. (10.8) to be

$$H(z) = g \frac{1 + \beta_1 z^{-1}+ \beta_2 z^{-2}}{1 + a_1 z^{-1}+ a_2 z^{-2}} \isdef \frac{B(z)}{A(z)}.$$

To specialize this to a second-order unity-gain allpass filter, we require

$$\left\vert H(e^{j\omega T})\right\vert = 1.$$

It is easy to show that, given any monic denominator polynomial $A(z)$, the numerator $B(z)$ must be, in the real case,^11.3

$$B(z) = z^{-2}A(z^{-1}) = a_2 + a_1z^{-1}+ z^{-2}.$$

Thus, to obtain an allpass biquad section, the numerator polynomial is simply the ``flip'' of the denominator polynomial. To obtain unity gain, we set $g=a_2$, $\beta_1 = a_1/a_2$, and $\beta_2=1/a_2$.

In terms of the poles and zeros of a filter $H(z)=B(z)/A(z)$, an allpass filter must have a zero at $z=1/p$ for each pole at $z=p$. That is if the denominator $A(z)$ satisfies $A(p)=0$, then the numerator polynomial $B(z)$ must satisfy $B(1/p)=0$. (Show this in the one-pole case.) Therefore, defining $B(z) = A(1/z)$ takes care of this property for all roots of $A(z)$ (all poles). However, since we prefer that $B(z)$ be a polynomial in $z^{-1}$, we define $B(z) = z^{-N}A(1/z)$, where $N$ is the order of $A(z)$ (the number of poles). $B(z)$ is then the flip of $A(z)$.

For further discussion and examples of allpass filters (including muli-input, multi-output allpass filters), see Appendix D.