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Embedded SystemsFPGAElectronics

Physical Audio Signal Processing
    Wave Digital Filters
       Wave Digital Elements
          Wave Digital Spring

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Wave Digital Spring

In the case of a spring with stiffness $ k$, we have the impedance

$\displaystyle R(s) = k/s $

which gives the reflectance

$\displaystyle S_k(s) = \frac{k/s - R_0}{k/s + R_0} $

As before, we may eliminate $ k$ by choosing $ R_0=k$ to get

$\displaystyle S_k(s) = \frac{1 - s }{1 + s} = z^{-1} $

under the bilinear transform. So we have the digital reflectance

$\displaystyle \fbox{$\displaystyle \tilde{S}_k(z) = z^{-1}$} \qquad\makebox[0pt][l]{(Wave Digital Spring)} $

and corresponding difference equation

$\displaystyle f^{{-}}(n) = f^{{+}}(n-1). $

Again the delay-free path has been eliminated. The wave flow diagram is shown in Fig.Q.3.

Figure Q.3: Wave flow diagram for the Wave Digital Spring.
\includegraphics{eps/lWaveDigitalSpring}

Thus, the WDF of a spring is simply a unit-sample delay, which is just the negative of the WDF mass. If we were to switch to velocity waves instead of force waves, both masses and springs would again correspond to unit-sample delays, but the spring would become inverting and the mass non-inverting.

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Previous: Wave Digital Mass
Next: Wave Digital Dashpot
written by 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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