Application of the Bilinear Transform

The impedance of a mass in the frequency domain is

$$R_M(s) = Ms.$$

In the $s$ plane, we have

$$F_a(s) = (Ms) V_a(s)$$

where the ``a'' subscript denotes ``analog''. For simplicity, let's choose the free constant $c$ in the bilinear transform such that $1$ rad/sec maps to one fourth the sampling rate, i.e., $s=j$ maps to $z=j$ which implies $c=1$ . Then the impedance relation maps across as

$$F_d(z) = \left(M\frac{1-z^{-1}}{1+z^{-1}}\right) V_d(z)$$

where the ``d'' subscript denotes ``digital. Multiplying through by the denominator and applying the shift theorem for $z$ transforms gives the corresponding difference equation

$$\begin{eqnarray*} (1+z^{-1})F_d(z) &=& M (1-z^{-1}) V_d(z) \\ \;\longleftrightarrow\;f_d(n) + f_d(n-1) &=& M[v_d(n) - v_d(n-1)] \\ \,\,\Rightarrow\,\,f_d(n) &=& M[v_d(n) - v_d(n-1)] - f_d(n-1). \end{eqnarray*}$$

This difference equation is diagrammed in Fig. 7.16. We recognize this recursive digital filter as the direct form I structure. The direct-form II structure is obtained by commuting the feedforward and feedback portions and noting that the two delay elements contain the same value and can therefore be shared [452]. The two other major filter-section forms are obtained by transposing the two direct forms by exchanging the input and output, and reversing all arrows. (This is a special case of Mason's Gain Formula which works for the single-input, single-output case.) When a filter structure is transposed, its summers become branching nodes and vice versa. Further discussion of the four basic filter section forms can be found in [336].

Figure 7.16: A direct-form-I digital filter simulating a mass $M$ created using the bilinear transform $s=(1-z^{-1})/(1+z^{-1})$ .