Simplified Impedance Analysis

The above results are quickly derived from the general reflection-coefficient for force waves (or voltage waves, pressure waves, etc.):

$$\zbox {\rho = \frac{R_2-R_1}{R_2+R_1} = \frac{\mbox{Impedance Step}}{\mbox{Impedance Sum}}} \protect$$ (5.22)

where $\rho$ is the reflection coefficient of impedance $R_2$ as ``seen'' from impedance $R_1$. In other words, if a wave $f^{{+}}$ traveling along in impedance $R_1$ suddenly hits a new impedance $R_2$, the wave will split into a reflected wave $f^{{-}}=\rho f^{{+}}$, and a transmitted wave $(1+\rho)f^{{+}}$. It therefore follows that a velocity wave $v^{+}$ will split into a reflected wave $v^{-}= - \rho v^{+}$ and transmitted wave $(1-\rho)v^{+}$. This rule is derived in §G.8.4 (and implicitly above as well).

In the mass-string-collision problem, we can immediately write down the force reflectance of the mass as seen from either string:

$$\hat{\rho}(s) = \frac{(ms+R) - R}{(ms+R) + R} = \frac{ms}{ms+2R}$$

That is, waves in the string are traveling through wave impedance $R$, and when they hit the mass, they are hitting the series combination of the mass impedance $R_m(s)=ms$ and the wave impedance $R$ of the string on the other side of the mass. Thus, in terms of Eq.$\,$(4.22) above, $R_1=R$ and $R_2=ms+R$.

The velocity reflectance is simply $-\hat{\rho}(s)$, since

$$\hat{\rho}(s) \isdef \frac{F^{-}}{F^{+}} = \frac{-RV^{-}}{RV^{+}} = - \frac{V^{-}}{V^{+}}.$$

The general rule for a force transmission-transfer-function (or force transmittance) is similarly easily derived:

$$\hat{\tau}_f(s) \isdef \frac{F}{F^{+}} = \frac{F^{+}+F^{-}}{F^{+}} = 1 + \frac{F^{-}}{F^{+}} = 1+\hat{\rho}(s)$$

A similar derivation shows that the velocity transmittance is also one plus the velocity reflectance:

$$\hat{\tau}_v(s) \isdef \frac{V}{V^{+}} = \frac{V^{+}+V^{-}}{V^{+}} = 1 + \frac{V^{-}}{V^{+}} = 1-\hat{\rho}(s)$$

That is, the transmittance is always one plus the reflectance, and the reflectance is $\hat{\rho}(s)$ for force waves and $-\hat{\rho}(s)$ for velocity waves. From the foregoing, we can immediately write down the force transmittance of a mass on a string:

$$\hat{\tau}_f(s) = 1+\hat{\rho}(s) = 1 + \frac{ms}{ms+2R} = 2\frac{ms+R}{ms+2R}$$

The velocity transmittance is

$$\hat{\tau}_v(s) = 1-\hat{\rho}(s) = 1 - \frac{ms}{ms+2R} = \frac{2R}{ms+2R}.$$

For $m=0$ the transmission filters become 1, as expected, since there is no mass on the string after all, and the wave simply propagates undisturbed to the other string segment. For $m=\infty$, $\hat{\tau}_f(s)\to 2$ while $\hat{\tau}_v(s)\to0$. The velocity transmittance $\hat{\tau}_v=0$ makes good physical sense because an infinite mass is a rigid termination of the string. Since the mass cannot be moved, the transmitted velocity must be zero. The fact that the transmitted force is twice the incoming force might give one pause; however, it cannot perform work on the other string, because the transmitted power is given by the force times the velocity, and the transmitted velocity is zero.

In summary, we have characterized the mass on the string in terms of its reflectance and transmittance from either string. For force waves, we have outgoing waves given by

$$\begin{eqnarray*} F^{-}_1(s) &=& \hat{\rho}(s) F^{+}_1(s) + \hat{\tau}_f(s) F^{-... ...}_2(s) &=& \hat{\tau}_f(s) F^{+}_1(s) + \hat{\rho}(s) F^{-}_2(s) \end{eqnarray*}$$

in terms of the incoming waves $F^{+}_1$ and $F^{-}_2$, the force reflectance $\hat{\rho}(s)=ms/(ms+2R)$, and the force transmittance $\hat{\tau}_f(s)=1+\hat{\rho}(s)=2(ms+R)/(ms+2R)$. We may say that the mass creates a dynamic scattering junction on the string. (If there were no dependency on $s$, such as when a dashpot is affixed to the string, we would simply call it a scattering junction.) The above form of the dynamic scattering junction is analogous to the Kelly-Lochbaum scattering junction (§G.8.4). The general relation $\hat{\tau}_f = 1+\hat{\rho}$ can be used to simplify the Kelly-Lochbaum form to a one-filter dynamic scattering junction analogous to the one-multiply scattering junction (§G.8.5):

$$\begin{eqnarray*} F^{-}_1 &=& \hat{\rho}F^{+}_1 + (1+\hat{\rho}) F^{-}_2 \;=\; F... ...at{\rho}F^{-}_2 \;=\; F^{+}_1 + \hat{\rho}\cdot(F^{+}_1+F^{-}_2) \end{eqnarray*}$$

The one-filter form follows from the observation that $\hat{\rho}\cdot(F^{+}_1+F^{-}_2)$ appears in both computations, and therefore need only be implemented once:

$$\begin{eqnarray*} F^{+}&\isdef & \hat{\rho}\cdot(F^{+}_1+F^{-}_2)\\ [5pt] F^{-}_... ...(1+\hat{\rho}) F^{+}_1 + \hat{\rho}F^{-}_2 \;=\; F^{+}_1 + F^{+} \end{eqnarray*}$$

This structure is diagrammed in Fig.4.23.

Figure 4.23: Continuous-time simulation diagram for for an ideal string with a point mass attached.

Again, the above results follow immediately from the more general formulation of §G.10.