DW Displacement Inputs

We define general DW inputs as follows:

$\displaystyle y^{+}_{n,m}$ $\displaystyle =$ $\displaystyle y^{+}_{n-1,m-1} + (\underline{\gamma}^{+}_m)^T \underline{\upsilon}(n)$ (E.33)
$\displaystyle y^{-}_{n,m}$ $\displaystyle =$ $\displaystyle y^{-}_{n-1,m+1} + (\underline{\gamma}^{-}_m)^T \underline{\upsilon}(n)$ (E.34)

The $ m$th $ 2q\times2$ block of the input matrix $ {\mathbf{B}_W}$ driving state components $ [y^{+}_{n+2,m},y^{-}_{n+2,m}]^T$ and multiplying $ [\underline{\upsilon}(n+2)^T,\underline{\upsilon}(n+1)^T]^T$ is then given by

$\displaystyle \left({\mathbf{B}_W}\right)_m = \left[\! \begin{array}{cc} (\unde...
...ma}^{-}_m)^T & (\underline{\gamma}^{-}_{m+1})^T \end{array} \!\right]. \protect$ (E.35)

Typically, input signals are injected equally to the left and right along the string, in which case

$\displaystyle \underline{\gamma}^{+}_m = \underline{\gamma}^{-}_m \isdef \underline{\gamma}_m.

Physically, this corresponds to applied forces at a single, non-moving, string position over time. The state update with this simplification appears as


Note that if there are no inputs driving the adjacent subgrid ( $ \underline{\gamma}_{m-1}=\underline{\gamma}_{m+1}=0$), such as in a half-rate staggered grid scheme, the input reduces to

\underline{x}_W(n+2) = \mathbf{A}_W\underline{x}_W(n) +

To show that the directly obtained FDTD and DW state-space models correspond to the same dynamic system, it remains to verify that $ \mathbf{A}_W=\mathbf{T}^{-1}\mathbf{A}_K\,\mathbf{T}$. It is somewhat easier to show that

\mathbf{T}\,\mathbf{A}_W&=& \mathbf{A}_K\,\mathbf{T}\\
...dots & \vdots & \vdots & \vdots & \vdots

A straightforward calculation verifies that the above identity holds, as expected. One can similarly verify $ \mathbf{C}_W=\mathbf{C}_K\,\mathbf{T}$, as expected. The relation $ {\mathbf{B}_W}=\mathbf{T}^{-1}\,\mathbf{B}_K$ provides a recipe for translating any choice of input signals for the FDTD model to equivalent inputs for the DW model, or vice versa. For example, in the scalar input case ($ q=1$), the DW input-weights $ {\mathbf{B}_W}$ become FDTD input-weights $ \mathbf{B}_K$ according to


The left- and right-going input-weight superscripts indicate the origin of each coefficient. Setting $ \gamma^{+}_m=\gamma^{-}_m$ results in

$\displaystyle \mathbf{B}_K= \left[\! \begin{array}{cc} \vdots & \vdots\\ \gamma...
...ma _{m+1}+\gamma _{m+3} \\ [5pt] \vdots & \vdots \end{array} \!\right] \protect$ (E.36)

Finally, when $ \gamma _m=1$ and $ \gamma _{\mu}=0$ for all $ \mu\neq m$, we obtain the result familiar from Eq.$ \,$(E.23):

\vdots & \vdots\\
...0 \\
2 & 0 \\
1 & 0 \\
\vdots & \vdots

Similarly, setting $ \gamma^{\pm}_{\mu}=0$ for all $ \mu\neq m+1$, the weighting pattern $ (1,2,1)$ appears in the second column, shifted down one row. Thus, $ \mathbf{B}_K$ in general (for physically stationary displacement inputs) can be seen as the superposition of weight patterns $ (1,2,1)$ in the left column for even $ m$, and the right column for odd $ m$ (the other subgrid), where the $ 2$ is aligned with the driven sample. This is the general collection of displacement inputs.

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DW Non-Displacement Inputs
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Proof that the Third-Order Time Derivative is Ill Posed