DW Displacement Inputs

We define general DW inputs as follows:

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

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

$$\left({\mathbf{B}_W}\right)_m = \left[\! \begin{array}{cc} (\underline{\gamma}^{+}_m)^T & (\underline{\gamma}^{+}_{m-1})^T \\ [5pt] (\underline{\gamma}^{-}_m)^T & (\underline{\gamma}^{-}_{m+1})^T \end{array} \!\right]. \protect$$ (E.35)

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

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

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

$$\underbrace{\left[\! \begin{array}{c} \vdots\\ y^{+}_{n+2,m}\\ [5pt] y^{-}_{n+2,m}\\ \vdots \end{array}\!\right]}_{\underline{x}_W(n+2)} =\mathbf{A}_W\underline{x}_W(n) + \underbrace{\left[\! \begin{array}{cc} \vdots & \vdots\\ \underline{\gamma}_m^T & \underline{\gamma}_{m-1}^T \\ [5pt] \underline{\gamma}_m^T & \underline{\gamma}_{m+1}^T \\ [5pt] \vdots & \vdots \end{array}\!\right]}_{{\mathbf{B}_W}} \underbrace{\left[\! \begin{array}{c} \underline{\upsilon}(n+2)\\ \underline{\upsilon}(n+1) \end{array}\!\right]}_{\underline{u}(n+2)}.$$

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

$$\underline{x}_W(n+2) = \mathbf{A}_W\underline{x}_W(n) + \underbrace{\left[\! \begin{array}{c} \vdots\\ \underline{\gamma}_{m-2}^T \\ [5pt] \underline{\gamma}_{m-2}^T \\ [5pt] \underline{\gamma}_m^T \\ [5pt] \underline{\gamma}_m^T \\ [5pt] \underline{\gamma}_{m+2}^T \\ [5pt] \underline{\gamma}_{m+2}^T \\ [5pt] \vdots \end{array}\!\right]}_{{\mathbf{B}_W}} \underline{\upsilon}(n+2).$$

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

$$\begin{eqnarray*} \mathbf{T}\,\mathbf{A}_W&=& \mathbf{A}_K\,\mathbf{T}\\ &=& \left[\! \begin{array}{cccccccccccc} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ \cdots & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & \cdots\\ & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{array}\!\right]. \end{eqnarray*}$$

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

$$\left[\! \begin{array}{l} \qquad\vdots\\ y_{n+1,m-1}\\ y_{n+2,m}\\ y_{n+1,m+1}\\ y_{n+2,m+2}\\ \qquad\vdots \end{array}\!\right] \! \leftarrow \! \underbrace{\left[\! \begin{array}{cc} \vdots & \vdots\\ \gamma^{+}_m +\gamma^{-}_{m-1} \,&\, \gamma^{+}_{m-1}+\gamma^{-}_{m-1} \\ [5pt] \gamma^{+}_m +\gamma^{-}_m \,&\, \gamma^{+}_{m-1}+\gamma^{-}_{m+1} \\ [5pt] \gamma^{-}_m +\gamma^{+}_{m+1} \,&\, \gamma^{+}_{m+1}\gamma^{-}_{m+1} \\ [5pt] \gamma^{+}_{m+2}+\gamma^{-}_{m+2} \,&\, \gamma^{+}_{m+1}+\gamma^{-}_{m+3} \\ [5pt] \vdots & \vdots \end{array}\!\right]}_{\mathbf{B}_K} \! \left[\! \begin{array}{c} \underline{\upsilon}(n+2)\\ \underline{\upsilon}(n+1) \end{array}\!\right]$$

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

$$\mathbf{B}_K= \left[\! \begin{array}{cc} \vdots & \vdots\\ \gamma _m +\gamma _{m-1} \,&\, 2\gamma _{m-1} \\ [5pt] 2\gamma _m \,&\, \gamma _{m-1}+\gamma _{m+1} \\ [5pt] \gamma _m +\gamma _{m+1} \,&\, 2\gamma _{m+1} \\ [5pt] 2\gamma _{m+2} \,&\, \gamma _{m+1}+\gamma _{m+3} \\ [5pt] \vdots & \vdots \end{array} \!\right] \protect$$ (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):

$$\mathbf{B}_K= \left[\! \begin{array}{cc} \vdots & \vdots\\ 1 & 0 \\ 2 & 0 \\ 1 & 0 \\ \vdots & \vdots \end{array}\!\right]$$

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.