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Properties of the Bilinear Transform

The bilinear transform maps an $ s$ -plane transfer function $ H_a(s)$ to a $ z$ -plane transfer function:

$\displaystyle \zbox{H_d(z) \;\mathrel{\stackrel{\mathrm{\Delta}}{=}}\;H_a\left(c\frac{1-z^{-1}}{1+z^{-1}}\right)}
$

We can observe the following properties of the bilinear transform:

Bilinear Transform of Force-Driven Mass

We have, from $ f=m\dot v\;\leftrightarrow\;F(s)=ms\,V(s)$ ,

$\displaystyle V(s) = \frac{1}{ms}F(s)
$

Setting $ s=(2/T)(1-z^{-1})/(1+z^{-1})$ according to the bilinear transform yields

$\displaystyle V_d(z) = \frac{T}{2m}\frac{1+z^{-1}}{1-z^{-1}}F_d(z)
$

where we defined

\begin{eqnarray*}
F_d(z) &=& F\left(\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}\right)\\ [10pt]
V_d(z) &=& V\left(\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}\right)
\end{eqnarray*}

The resulting finite-difference scheme is then

$\displaystyle v_d(n) - v_d(n-1) = \frac{T}{2m}\left[f_d(n)+f_d(n-1)\right]
$

i.e.,

$\displaystyle v_d(n) = v_d(n-1) + \frac{T}{2m}\left[f_d(n)+f_d(n-1)\right]
$

We see that this is the same as the backward Euler scheme plus a new term $ (T/2m) f_d(n-1)$ .


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``Introduction to Physical Signal Models'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2020-06-27 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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