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Simplified Impedance Analysis

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

$\displaystyle \zbox {\rho = \frac{R_2-R_1}{R_2+R_1} = \frac{\mbox{Impedance Step}}{\mbox{Impedance Sum}}} \protect$ (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 §H.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:

$\displaystyle \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

$\displaystyle \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:

$\displaystyle \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: