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Embedded SystemsFPGAElectronics

Physical Audio Signal Processing
    Virtual Musical Instruments
       Woodwinds
          Single-Reed Theory
             Scattering-Theoretic Formulation

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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$.

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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