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Embedded SystemsFPGAElectronics

Physical Audio Signal Processing
    Digital Waveguide Theory
       Non-Cylindrical Acoustic Tubes
          Momentum Conservation in Nonuniform Tubes

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Momentum Conservation in Nonuniform Tubes

Newton's second law ``force equals mass times acceleration'' implies that the pressure gradient in a gas is proportional to the acceleration of a differential volume element in the gas. Let $ A(x)$ denote the area of the surface of constant phase at radial coordinate $ x$ in the tube. Then the total force acting on the surface due to pressure is $ f(t,x)=A(x)p(t,x)$, as shown in Fig.C.45.

Figure C.45: Differential volume element for the conical acoustic tube.
\includegraphics[width=3in]{eps/fconic}

The net force $ df(t,x)$ to the right across the volume element between $ x$ and $ x+dx$ is then

$\displaystyle df(t,x) = f(t,x)-f(t,x+dx)$ $\displaystyle =$ $\displaystyle f(t,x)-[f(t,x) + dx \cdot f'(x)]$  
  $\displaystyle =$ $\displaystyle - dx \cdot f'(x)$  
  $\displaystyle =$ $\displaystyle - dx \cdot[A(x)p(t,x)]'$  
  $\displaystyle =$ $\displaystyle -dx \cdot[A' p + A'p],$  

where, when time and/or position arguments have been dropped, as in the last line above, they are all understood to be $ t$ and $ x$, respectively. To apply Newton's second law equating net force to mass times acceleration, we need the mass of the volume element

\begin{eqnarray*} dM(t,x) &=& \int_x^{x+dx} \rho(t,\xi) A(\xi)\,d\xi \\ [5pt] &\... ...\rho A' \right)\frac{(dx)^2}{2}\\ [5pt] &\approx& \rho\, A\,dx, \end{eqnarray*}

where $ \rho(t,x)$ denotes air density.

The center-of-mass acceleration of the volume element can be written as $ {\dot u}(t,x)$ where $ u(t,x)$ is particle velocity.C.16 Applying Newton's second law $ df = dM\cdot {\dot u}$, we obtain

$\displaystyle -dx \cdot (A'p + Ap') \eqsp \rho\, A\,dx\, {\dot u} \protect$ (C.146)

or, dividing through by $ -A\,dx$,

$\displaystyle p' + p \, \frac{A'}{A} \eqsp - \rho\,{\dot u}. \protect$ (C.147)

In terms of the logarithmic derivative of $ A$, this can be written

$\displaystyle p' + p \,$   ln$\displaystyle 'A \eqsp - \rho\,{\dot u}. \protect$ (C.148)

Note that $ p$ denotes small-signal acoustic pressure, while $ \rho$ denotes the full gas density (not just an acoustic perturbation in the density). We may therefore treat $ \rho$ as a constant.


Subsections
Previous: Digital Simulation
Next: Cylindrical Tubes

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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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