Finite Difference Approximation

As introduced in Appendix M, the finite difference approximation (FDA) amounts to replacing derivatives by finite differences, or

$$\frac{d}{dt} x(t) \isdef \lim_{\delta\to 0} \frac{x(t) - x(t-\delta)}{\delta} \approx \frac{x(n T)-x[(n-1)T]}{T} \protect$$ (K.2)

for sufficiently small $T$.^K.5

See §G.2.1 for a discussion of using the FDA to model ideal vibrating strings.

Viewing Eq.$\,$(K.2) in the frequency domain, the transfer function of an ideal differentiator is $H(s)=s$, which can be viewed as the Laplace transform of the operator $d/dt$ (left-hand side of Eq.$\,$(K.2)). Moving to the right-hand side, the z transform of the first-order difference operator is $(1-z^{-1})/T$. Thus, in the frequency domain, the finite-difference approximation may be performed by making the substitution

$$s \leftarrow \frac{1-z^{-1}}{T} \protect$$ (K.3)

in any continuous-time transfer function (Laplace transform of an integro-differential operator) to obtain a discrete-time transfer function (z transform of a finite-difference operator).

The inverse of substitution Eq.$\,$(K.3) is

$$z = \frac{1}{1 - sT} = 1 + sT+ (sT)^2 + \cdots \, .$$

The FDA is a special case of the matched $z$ transformation [366] applied to the point $s=0$. In general, the matched $z$ transformation maps a pole at $s = -a$ to the point $z = e^{-aT}$, where $T$ is the sampling period. Thus, each pole and zero are mapped according to

$$z_i = e^{s_i T} = 1 + {s_i T} + \frac{(s_i T)^2}{2} + \cdots\, .$$

The actual transformation is carried out by factoring $H(s)$ into a product of first-order terms such as $(s+a)$, and substituting

$$s+a \to 1 - z^{-1}e^{-aT}.$$

Setting $a=0$ gives the FDA for $T=1$.^K.6

Since the FDA is the matched $z$ transformation for poles and zeros at the origin of the $s$ plane, it follows that it maps analog dc ($s=0$) to digital dc ($z=1$). However, that is the only ideal mapping in the frequency domain, as discussed further below.

Note that the FDA does not alias, since the conformal mapping $s = {1 - z^{-1}}$ is one to one, but it does warp the poles and zeros in a way which may not be desirable.

It is convenient to think of the FDA in terms of time-domain difference operators using a delay operator notation. The delay operator $d$ is defined by

$$d^k x(n) = x(n-k).$$

Thus, the first-order difference (derivative approximation) is represented in the time domain by $(1-d)/T$. We can think of $d$ as $z^{-1}$ since, by the shift theorem for $z$ transforms, $z^{-k} X(z)$ is the $z$ transform of $x$ delayed (right shifted) by $k$ samples.

The obvious definition for the second derivative is

$${\hat{\ddot x}}(n) = \frac{(1-d)^2}{T^2} x(n)$$ (K.4)

However, a better definition is

$${\hat{\ddot x}}(n) \isdef \frac{(d^{-1}-1)(1-d)}{T^2} x(n) = \frac{d^{-1}-2+d}{T^2}x(n)$$ (K.5)

where $d^{-1}$ denotes a unit-sample advance. This definition is preferable as long as one sample of look-ahead is available, since it avoids an operator delay of one sample. Equation (K.5) is a zero phase filter, meaning it has no delay at any frequency, while (K.4) is a linear phase filter having a delay of $1$ sample at all frequencies.