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Design 2: Bilinear Transform Method

As discussed in §L.4, the bilinear transform maps the $ s$ plane to the $ z$ plane via a first-order conformal map. Specifically, the analog transfer function $ H(s)$ is mapped to the $ z$ plane by means of the change of variable

$\displaystyle s = \alpha \frac{1-z^{-1}}{1+z^{-1}}. \protect$ (D.5)

Carrying this out on Eq.$ \,$(D.3) yields

$\displaystyle H_{L,\omega_1}(z) = \frac{\omega_1}{\alpha+\omega_1}\frac{1+z^{-1}}{1-pz^{-1}} \protect$ (D.6)

where the pole $ p$ is given by

$\displaystyle p = \frac{\alpha - \omega_1}{\alpha+\omega_1}.
$

The real constant $ \alpha>0$ may be chosen to map any $ s$-plane frequency exactly to any $ z$-plane frequency. In our case, we clearly want $ \omega_1$ to be mapped exactly. Specializing Eq.$ \,$(D.5) to the analog and digital frequency axes gives

\begin{eqnarray*}
j\omega_a
&=& \alpha \frac{1-e^{-j\omega_d T}}{1+e^{-j\omega_...
...2)}{2\cos(\omega_d T/2)}\\ [10pt]
&=& j\alpha \tan(\omega_d T/2)
\end{eqnarray*}

or

$\displaystyle \zbox {\omega_a = \alpha \tan\left(\frac{\omega_d T}{2}\right).}
$

Thus, to exactly map the behavior at $ \omega_a=\omega_1$ to $ \omega_d=\omega_1$, we may set $ \alpha$ to

$\displaystyle \zbox {\alpha = \frac{\omega_1}{\tan\left(\frac{\omega_1 T}{2}\right)}.}
$

However, note that if the