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Centered Finite Difference

Backward Euler [ $ s\leftarrow (1-z^{-1})/T)$ ] has a 1/2 sample delay at all frequencies, while Forward Euler [ $ s\leftarrow(z-1)/T)$ ] has a 1/2 sample advance. We can eliminate this time-skew using a
centered finite difference:

\begin{eqnarray*}
\hat{\dot v}(nT) &=& \frac{v_{n+1}-v_{n-1}}{2T}\\ [10pt]
\Rightarrow\quad f_{n}&\approx&\frac{m}{2T}(v_{n+1}-v_{n-1})\\ [10pt]
\Rightarrow\quad \hat{v}_{n+1} &=& \hat{v}_{n-1} + \frac{2T}{m}f_{n}
\end{eqnarray*}


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``Introduction to Physical Signal Models'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2020-06-27 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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