A More Formal Derivation of the Wave Digital Force-Driven Mass

Above we derived how to handle the external force by direct physical reasoning. In this section, we'll derive it using a more general step-by-step procedure which can be applied systematically to more complicated situations.

Figure F.12 gives the physical picture of a free mass driven by an external force in one dimension. Figure F.13 shows the electrical equivalent circuit for this scenario in which the external force is represented by a voltage source emitting $f(t)$ volts, and the mass is modeled by an inductor having the value $L=m$ Henrys.

Figure F.12: Physical diagram of an external force driving a mass sliding on a frictionless surface.

Figure F.13: Electrical equivalent circuit of the force-driven mass in Fig.F.12.

The next step is to convert the voltages and currents in the electrical equivalent circuit to wave variables. Figure F.14 gives an intermediate equivalent circuit in which an infinitesimal transmission line section with real impedance $R_0 = m$ has been inserted to facilitate the computation of the wave-variable reflectance, as we did in §F.1.1 to derive Eq.(F.1).

Figure F.14: Intermediate equivalent circuit for the force-driven mass in which an infinitesimal transmission line section has been inserted to facilitate conversion of the mass impedance $ms$ into a wave-variable reflectance.

Figure F.15: Intermediate wave-variable model of the force-driven mass of Fig.F.13.

Figure F.15 depicts a next intermediate equivalent circuit in which the mass has been replaced by its reflectance (using ``$z^{-1}$ '' to denote the continuous-time reflectance $(1-s)/(1+s)$ , as derived in §F.1.1). The infinitesimal transmission-line section is now represented by a ``resistor'' since, when the voltage source is initially ``switched on'', it only ``sees'' a real resistance having the value $m$ Ohms (the waveguide interface). After a short period of time determined by the reflectance of the mass,^F.3 ``return waves'' from the mass result in an ultimately reactive impedance. This of course must be the case because the mass does not dissipate energy. Therefore, the ``resistor'' of $m$ Ohms is not a resistor in the usual sense since it does not convert potential energy (the voltage drop across it) into heat. Instead, it converts potential energy into propagating waves with 100% efficiency. Since all of this wave energy is ultimately reflected by the terminating element (mass, spring, or any combination of masses and springs), the net effect is a purely reactive impedance, as we know it must be.

Figure F.16: Interconnection of the wave digital mass with an ideal force source by means of a two-port parallel adaptor. The symbol ``$\vert\vert$ '' is used in the WDF literature to signify a parallel adaptor.

To complete the wave digital model, we need to connect our wave digital mass to an ideal force source which asserts the value $f(n)$ each sample time. Since an ideal force source has a zero internal impedance, we desire a parallel two-port junction which connects the impedances $R_1 = 0$ ( $\Gamma _1=\infty$ ) and $R_2=m$ ( $\Gamma _2=1/m$ ), as shown in Fig.F.16. From Eq.(F.15) we have that the common junction force is equal to

$$\begin{eqnarray*} f(n) &=& \alpha_1 f^{{+}}_1(n) + \alpha_2 f^{{+}}_2(n)\\ &=& \frac{2\Gamma _1}{\Gamma _1+\Gamma _2} f^{{+}}_1(n) + \frac{2\Gamma _2}{\Gamma _1+\Gamma _2} f^{{+}}_2(n)\\ &=& \frac{2\cdot\infty}{\infty+\frac{1}{m}} f^{{+}}_1(n) + \frac{\frac{2}{m}}{\infty+\frac{1}{m}} f^{{+}}_2(n)\\ &=& 2 f^{{+}}_1(n) \end{eqnarray*}$$

from which we conclude that

$$\fbox{$\displaystyle f^{{+}}_1(n) = \frac{f(n)}{2}$}$$

The outgoing waves are, by Eq.(F.16),

$$\begin{eqnarray*} f^{{-}}_1(n) &=& f(n) - f^{{+}}_1(n) = \frac{f(n)}{2}\\ f^{{-}}_2(n) &=& f(n) - f^{{+}}_2(n) = f(n) - f^{{-}}_m(n) \end{eqnarray*}$$

Since $\alpha_1=2$ and $\alpha_2=0$ for this model, the reflection coefficient seen on port 1 is $\rho = \alpha_1 - 1 = 1$ . The transmission coefficient from port 1 is $1+\rho=2$ . In the opposite direction, the reflection coefficient on port 2 is $-\rho= -1$ , and the transmission coefficient from port 2 is $1-\rho = 0$ . The final result, drawn in Kelly-Lochbaum form (see §F.2.1), is diagrammed in Fig.F.17, as well as the result of some elementary simplifications. The final model is the same as in Fig.F.11, as it should be.

Figure F.17: Wave digital mass driven by external force $f(n)$ .