Simplified Impedance Analysis

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

$$\zbox {\rho = \frac{R_2-R_1}{R_2+R_1} = \frac{\mbox{Impedance Step}}{\mbox{Impedance Sum}}} \protect$$ (10.17)

where $\rho$ is the reflection coefficient of impedance $R_2$ as ``seen'' from impedance $R_1$ . If a force 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 §C.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:

$$\zbox {\hat{\rho}_f(s) \eqsp \frac{(ms+R) - R}{(ms+R) + R} \eqsp \frac{ms}{ms+2R}}$$

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 $ms$ and the wave impedance $R$ of the string on the other side of the mass. Thus, in terms of Eq.(9.17) above, $R_1=R$ and $R_2=ms+R$ .

Since, by the Ohm's-law relations,

$$\hat{\rho}_f(s) \isdefs \frac{F^{-}}{F^{+}} \eqsp \frac{-RV^{-}}{RV^{+}} \eqsp - \frac{V^{-}}{V^{+}} \isdefs -\hat{\rho}_v(s).$$

we have that the velocity reflectance is simply

$$\zbox {\hat{\rho}_v(s) \isdefs \frac{V^{-}}{V^{+}} \eqsp -\hat{\rho}_f(s) \isdefs -\frac{F^{-}}{F^{+}}.}$$