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As Recursion

By expanding out the behaviour of the operator $ \delta _{tt}$, and reintroducing the time index $ n$, one loses compactness of representation, but has a clearer picture of how to program such a scheme on a computer. One has

$\displaystyle \frac{1}{k^2}\left(u^{n+1}-2u^{n}+u^{n-1}\right) = -\omega_{0}^2 u^n$ (3.19)

which involves values of the time series at three levels $ n+1$, $ n$, and $ n-1$. Rewriting (3.19) so that $ u^{n+1}$ is isolated gives

$\displaystyle u^{n+1} = \left(2-\omega_{0}^2 k^2\right) u^n-u^{n-1}$ (3.20)

which is a two-step recursion in the time series $ u^{n}$. This difference equation is identical in form to a two-pole digital filter under zero-input conditions.



Stefan Bilbao 2006-11-15