Scattering-Theoretic Formulation

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

$$p_b^{-} = \frac{1-r}{1+r}p_b^{+}+ \frac{r}{1+r}p_m$$ (10.41)

$$= \rho p_b^{+}+ \frac{1-\rho}{2} p_m$$ (10.42)

$$= \frac{p_m}{2} - \rho \frac{p_{\Delta}^{+}}{2}$$ (10.43)

where

$$\rho(p_{\Delta}) \isdef \frac{1-r(p_{\Delta})}{1+r(p_{\Delta})}, \qquad r(p_{\Delta})\isdef \frac{R_b}{R_m(p_{\Delta})}$$ (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 $p_b^{-}= -p_b^{+}+ p_m$ .