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

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

This difference equation is diagrammed in Fig. J.11. 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. 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 [312].

Figure J.11: A direct-form-I digital filter simulating a mass $M$ created using the bilinear transform $s=(1-z^{-1})/(1+z^{-1})$.