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Digital Waveguide (DW) Scheme

We now derive the digital waveguide formulation by sampling the traveling-wave solution to the wave equation. It is easily checked that the lossless 1D wave equation $ Ky''=\epsilon {\ddot y}$ is solved by any string shape $ y$ which travels to the left or right with speed $ c \isdeftext \sqrt{K/\epsilon }$ [100]. Denote right-going traveling waves in general by $ y_r(t-x/c)$ and left-going traveling waves by $ y_l(t+x/c)$, where $ y_r$ and $ y_l$ are assumed twice-differentiable. Then, as is well known, the general class of solutions to the lossless, one-dimensional, second-order wave equation can be expressed as

$\displaystyle y(t,x) = y_r\left(t-\frac{x}{c}\right) + y_l\left(t+\frac{x}{c}\right). \protect$ (E.4)

Sampling these traveling-wave solutions yields
$\displaystyle y(nT,mX)$ $\displaystyle =$ $\displaystyle y_r(nT-mX/c) + y_l(nT+mX/c)$  
  $\displaystyle =$ $\displaystyle y_r[(n-m)T] + y_l[(n+m)T]$  
  $\displaystyle \isdef$ $\displaystyle y^{+}(n-m) + y^{-}(n+m)
\protect$ (E.5)

where a ``$ +$'' superscript denotes a ``right-going'' traveling-wave component, and ``$ -$'' denotes propagation to the ``left''. This notation is similar to that used for acoustic-tube modeling of speech [297].

Figure E.1: Digital simulation of the ideal, lossless waveguide with observation points at $ x=0$ and $ x=3X=3cT$.
\includegraphics[width=\twidth]{eps/fidealCopy}

Figure E.1 (initially given as Fig.C.3) shows a signal flow diagram for the computational model of Eq.$ \,$(E.5), termed a digital waveguide model (developed in detail in Appendix C). Recall that, by the sampling theorem, it is an exact model so long as the initial conditions and any ongoing additive excitations are bandlimited to less than half the temporal sampling rate $ f_s=1/T$ [451, Appendix G]. Recall also that the position along the string, $ x_m = mX= m cT$ meters, is laid out from left to right in the diagram, giving a physical interpretation to the horizontal direction in the diagram, even though spatial samples have been eliminated from explicit consideration. (The arguments of $ y^{+}$ and $ y^{-}$ have physical units of time.)

The left- and right-going traveling wave components are summed to produce a physical output according to

$\displaystyle y(nT,mX) = y^{+}(n-m) + y^{-}(n+m) \protect$ (E.6)

In Fig.E.1, ``transverse displacement outputs'' have been arbitrarily placed at $ x=0$ and $ x=3X$.


Previous: Finite Difference Time Domain (FDTD) Scheme
Next: FDTD and DW Equivalence

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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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