Scattering-Theoretic Formulation

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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)

where

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

(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 goes to infinity relative to , (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, , then we would have 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 (regarding it now as a tube section of infinite radius), then $r=\infty$ , $\rho=-1$ , and $p_b^{-}= -p_b^{+}+ p_m$ .