Scattering-Theoretic Formulation

Equation (9.38) can be solved for $ p_b^{-}$ to obtain

$\displaystyle p_b^{-}$ $\displaystyle =$ $\displaystyle \frac{1-r}{1+r}p_b^{+}+ \frac{r}{1+r}p_m$ (10.41)
  $\displaystyle =$ $\displaystyle \rho p_b^{+}+ \frac{1-\rho}{2} p_m$ (10.42)
  $\displaystyle =$ $\displaystyle \frac{p_m}{2} - \rho \frac{p_{\Delta}^{+}}{2}$ (10.43)

$\displaystyle \rho(p_{\Delta}) \isdef \frac{1-r(p_{\Delta})}{1+r(p_{\Delta})}, \qquad r(p_{\Delta})\isdef \frac{R_b}{R_m(p_{\Delta})}$     (10.44)

We interpret $ \rho(p_{\Delta})$ as a signal-dependent reflection coefficient.

Since the mouthpiece of a clarinet is nearly closed, $ R_m\gg R_b$ which implies $ r\approx 0$ and $ \rho\approx1$. In the limit as $ R_m$ goes to infinity relative to $ R_b$, (9.42) reduces to the simple form of a rigidly capped acoustic tube, i.e., $ p_b^{-}= p_b^{+}$. If it were possible to open the reed wide enough to achieve matched impedance, $ R_m=R_b$, then we would have $ r=1$ and $ \rho=0$, in which case $ p_b^{-}= p_m/2$, with no reflection of $ p_b^{+}$, as expected. If the mouthpiece is removed altogether to give $ R_m=0$ (regarding it now as a tube section of infinite radius), then $ r=\infty$, $ \rho=-1$, and $ p_b^{-}=
-p_b^{+}+ p_m$.

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