The Bilinear Transform

The formula for a general first-order (bilinear) conformal mapping of functions of a complex variable is conveniently expressed by [3, page 75]

$${(\zeta -\zeta _1)(\zeta _2-\zeta _3) \over (\zeta _2-\zeta _1)(\zeta -\zeta _3)} = {(z-z_1)(z_2-z_3) \over (z_2-z_1)(z-z_3)}.$$ (2)

It can be seen that choosing three specific points and their images determines the mapping for all z and $\zeta$.

Bilinear transformations map circles and lines into circles and lines (lines being viewed as circles passing through the point at infinity). In digital audio, where both domains are ``z planes,'' we normally want to map the unit circle to itself, with dc mapping to dc ( $z_1=\zeta _1=1$) and half the sampling rate mapping to half the sampling rate ( $z_2=\zeta _2=-1$). Making these substitutions in Eq.(2) leaves us with transformations of the form

$$z= {\cal A}_{\rho }(\zeta ) = {\zeta + \rho \over 1 + \zeta \rho } , \qquad \rho = {\zeta _3 - z_3 \over 1 - z_3\zeta _3}.$$ (3)

The constant $\rho$ provides one remaining degree of freedom which can be used to map any particular frequency $\omega$ (corresponding to the point $e^{j\omega }$ on the unit circle) to a new location $a(\omega )$. All other frequencies will be warped accordingly. The allpass coefficient $\rho$ can be written in terms of these frequencies as

$$\rho = {\sin\{[a(\omega )-\omega ]/2\} \over \sin\{[a(\omega )+\omega ]/2\} },$$ (4)

In this form, it is clear that $\rho$ is real and that the inverse of ${\cal A}_{\rho }$ is ${\cal A}_{-\rho }$. Also, since $0\leq\{\omega ,a(\omega )\}\leq\pi$, and $a(\omega )\geq\omega$ for a Bark map, we have $\rho \in[0,1)$ for a Bark map from the z plane to the $\zeta$ plane.