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Power Waves

Physically,

\begin{eqnarray*}
\hbox{Power} &=& \hbox{Work} / \hbox{Time} \\
&=& \hbox{Force} \times \hbox{Distance} / \hbox{Time} \\
&=& \hbox{Force} \times \hbox{Velocity}
\end{eqnarray*}

Traveling power waves:

\begin{displaymath}
\begin{array}{rcrl}
{\cal P}^{+}(n) &\mathrel{\stackrel{\mathrm{\Delta}}{=}}&&f^{{+}}(n)v^{+}(n) \\
{\cal P}^{-}(n) &\mathrel{\stackrel{\mathrm{\Delta}}{=}}&-&f^{{-}}(n)v^{-}(n) \nonumber
\end{array}\end{displaymath}

From ``Ohm's law'' $ f^{{+}}= Rv^{+}$ and $ f^{{-}}= -Rv^{-}$ , we have

\begin{eqnarray*}
{\cal P}^{+}(n)&=&R\,[v^{+}(n)]^2=\frac{[f^{{+}}(n)]^2}{R}
\\
{\cal P}^{-}(n)&=&R\,[v^{-}(n)]^2=\frac{[f^{{-}}(n)]^2}{R} \nonumber
\end{eqnarray*}

Note that both $ {\cal P}^{+}$ and $ {\cal P}^{-}$ are nonnegative

Summing traveling powers gives total power:

$\displaystyle {\cal P}(t_n,x_m) \mathrel{\stackrel{\mathrm{\Delta}}{=}}{\cal P}^{+}(n-m) + {\cal P}^{-}(n+m)
$

If we had instead defined $ {\cal P}^{-}(n) \mathrel{\stackrel{\mathrm{\Delta}}{=}}f^{{-}}(n)v^{-}(n)$ (no minus sign in front), then summing the traveling powers would give net power flow.


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``Choice of Wave Variables in Digital Waveguide Models'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2019-02-05 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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