Finite Difference Approximation vs. 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), and this 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 the scale factor) 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 a potentially problematic revision of the differentiator. To get something more 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 §8.6) [365].

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

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

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

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

We see that analog dc ($s=0$ ) maps to digital 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

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

From this it can be checked that the FDA maps the $j\omega$ axis in the $s$ plane to the circle of radius $1/2$ centered at the point $z = 1/2$ in the $z$ plane, as shown in Fig. 7.15

Figure 7.15: Image of the $j\omega$ axis in the $z$ plane: a circle of radius $1/2$ centered at the point $z = 1/2$ .

Under the FDA, analog and digital frequency axes coincide well enough at very low frequencies (high sampling rates), but at high frequencies relative to the sampling rate, artificial damping is introduced as the image of the $j\omega$ axis diverges away from the unit circle.

While the bilinear transform ``warps'' the frequency axis, we can say the FDA ``doubly warps'' the frequency axis: It has a progressive, compressive warping in the direction of increasing frequency, like the bilinear transform, but unlike the bilinear transform, it also warps normal to the frequency axis.

Consider a point traversing the upper half of the unit circle in the z plane, starting at $z=1$ and ending at $z=-1$ . At dc, the FDA is perfect, but as we proceed out along the unit circle, we diverge from the $j\omega$ axis image and carve an arc somewhere out in the image of the right-half $s$ plane. This has the effect of introducing an artificial damping.

Consider, for example, an undamped mass-spring system. There will be a complex conjugate pair of poles on the $j\omega$ axis in the $s$ plane. After the FDA, those poles will be inside the unit circle, and therefore damped in the digital counterpart. The higher the resonance frequency, the larger the damping. It is even possible for unstable $s$ -plane poles to be mapped to stable $z$ -plane poles.

In summary, both the bilinear transform and the FDA preserve order, stability, and positive realness. They are both free of aliasing, high frequencies are compressively warped, and both become ideal at dc, or as $f_s$ approaches $\infty$ . However, at frequencies significantly above zero relative to the sampling rate, only the FDA introduces artificial damping. The bilinear transform maps the continuous-time frequency axis in the $s$ (the $j\omega$ axis) plane precisely to the discrete-time frequency axis in the $z$ plane (the unit circle).