Wave Digital Mass

In the case of a mass $m$ , we have

$$R(s) = m s$$   (Mass Impedance)

which implies its reflectance is, from Eq.(F.10),

$$\hat{\rho}_m(s) = \frac{m s - R_0 }{m s + R_0}$$

Setting $R_0 = m$ gives

$$\hat{\rho}_m(s) = \frac{s - 1 }{s + 1}$$

and this choice also turns out to eliminate the delay-free path in the digital version. In view of the expression for the inverse bilinear transform in Eq.(F.9), i.e., $z=(1+s)/(1-s)$ , the bilinear transform of $\hat {\rho }(s)$ is immediately seen to be

$$\fbox{$\displaystyle \hat{\tilde{\rho}}_m(z) = -z^{-1}$} \qquad\mbox{(Wave Digital Mass)}$$

where we defined $\hat{\tilde{\rho}}_m(z) \isdef \hat{\rho}_m\left(\frac{z-1}{z+1}\right)$ . The corresponding difference equation for the wave digital mass is

$$f^{{-}}(n) = - f^{{+}}(n-1)$$

and its wave flow diagram is drawn in Fig.F.2.

Figure F.2: Wave flow diagram for the wave digital mass. Note that the wave variables are written in the time domain as is customary in digital filter diagrams, while it would be more consistent (with the $z^{-1}$ block) to keep them in the frequency domain as $F^{+}(z)$ and $F^{-}(z)$ .

Thus, the wave digital mass is simply a unit-sample delay and a negation. The fact that the value of the mass has been canceled out will be addressed below in the subsection on ``adaptors,'' i.e., it only affects interconnection with other elements. For now, just remember that the reference impedance was chosen to be equal to the mass in order to get this simple wave flow diagram. Also note that the WDF mass simulator has no delay-free path from input to output.