Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

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)

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 $ 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$ .


Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

[How to cite this work]  [Order a printed hardcopy]  [Comment on this page via email]

``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA