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Physical Audio Signal Processing
    Introduction to Lumped Models
       Bilinear Transformation
          Finite Differences vs. the Bilinear Transform

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Finite Differences vs. the Bilinear Transform

Recall that the finite difference approximation (FDA) defines the elementary differentiator by $ y(n) = x(n) - x(n-1)$ (ignoring the scale factor $ T$ for now) which approximates the ideal transfer function $ H(s)=s$ by $ H_d(z)=1-z^{-1}$. The bilinear transform calls instead for the transfer function $ H'_d(z)=(1-z^{-1})/(1+z^{-1})$ (again dropping scale factors) which introduces a pole at $ z=-1$ and gives us the recursion $ y(n) = x(n) - x(n-1) - y(n-1)$. Note that this new pole is right on the unit circle and is therefore undamped. Any signal energy at half the sampling rate will circulate forever in the recursion, and due to round-off error, it will tend to grow. This is therefore not a very useful improvement of the differentiator. To get something really practical, we need to specify that the filter frequency response approximate $ H(j\omega)=j\omega$ over a finite range of frequencies $ [-\omega_c,\omega_c]$, where $ \omega_c\ll\pi f_s$, above which we allow the response to ``roll off'' to zero. This is how we pose the differentiator problem in terms of general purpose filter design (see §R.3) [371].

To understand the properties of the finite difference approximation in the frequency domain, we may look at the properties of its $ s$-plane to $ z$-plane mapping

$\displaystyle s = \frac{1 - z^{-1}}{T} $

We see the FDA is actually a portion of the bilinear transform, since following the FDA mapping by the mapping $ s = (c/T)/(1+z^{-1})$ would convert it to the bilinear transform. Like the bilinear transform, the FDA does not alias, since the mapping $ s = 1 - z^{-1}$ is one-to-one.

Setting $ T$ to 1 for simplicity and solving the FDA mapping for z gives

$\displaystyle z = \frac{1 }{1-s} $

We see that dc ($ s=0$) maps to dc ($ z=1$) as desired, but higher frequencies unfortunately map inside the unit circle rather than onto the unit circle in the $ z$ plane. Solving for the image in the z plane of the $ j\omega $ axis in the s plane gives

$\displaystyle z = \frac{1}{1-j \omega } = \frac{1 + j \omega }{1+\omega^2} $

From this it can be checked that the FDA maps the