FDTD State Space Model

Let $\underline{x}_K(n)$ denote the FDTD state for one of the two subgrids at time $n$ , as defined by Eq.(E.10). The other subgrid is handled identically and will not be considered explicitly. In fact, the other subgrid can be dropped altogether to obtain a half-rate, staggered grid scheme [55,148]. However, boundary conditions and input signals will couple the subgrids, in general. To land on the same subgrid after a state update, it is necessary to advance time by two samples instead of one. The state-space model for one subgrid of the FDTD model of the ideal string may then be written as

$$\underline{x}_K(n+2) = \mathbf{A}_K\, \underline{x}_K(n) + \mathbf{B}_K\, \underline{u}(n+2)$$

$$\underline{y}(n) = \mathbf{C}_K\, \underline{x}_K(n). \protect$$ (E.24)

To avoid the issue of boundary conditions for now, we will continue working with the infinitely long string. As a result, the state vector $\underline{x}_K(n)$ denotes a point in a space of countably infinite dimensionality. A proper treatment of this case would be in terms of operator theory [328]. However, matrix notation is also clear and will be used below. Boundary conditions are taken up in §E.4.3.

When there is a general input signal vector $\underline{u}(n)$ , it is necessary to augment the input matrix $\mathbf{B}_K$ to accomodate contributions over both time steps. This is because inputs to positions $m\pm1$ at time $n+1$ affect position $m$ at time $n+2$ . Henceforth, we assume $\mathbf{B}_K$ and $\underline{u}$ have been augmented in this way. Thus, if there are $q$ input signals $\underline{\upsilon}(n)\isdeftext [\upsilon _i(n)]$ , $i=1,\ldots,q$ , driving the full string state through weights $\underline{\beta}_m\isdeftext [\beta_{m,i}]$ , $m\in\mathbb{Z}$ , the vector $\underline{u}(n)=$ is of dimension $2q\times 1$ :

$$\underline{u}(n+2) = \left[\! \begin{array}{c} \underline{\upsilon}(n+2)\\ \underline{\upsilon}(n+1) \end{array}\!\right]$$

When there is only one physical input, as is typically assumed for plucked, struck, and bowed strings, then $q=1$ and $\underline{u}$ is $2\times 1$ . The matrix $\mathbf{B}_K$ weights these inputs before they are added to the state vector for time $n+2$ , and its entries are derived in terms of the $\beta_{m,i}$ coefficients below.

$\mathbf{C}_K$ forms the output signal as an arbitrary linear combination of states. To obtain the usual displacement output for the subgrid, $\mathbf{C}_K$ is the matrix formed from the identity matrix by deleting every other row, thereby retaining all displacement samples at time $n$ and discarding all displacement samples at time $n-1$ in the state vector $\underline{x}_K(n)$ :

$$\underbrace{\left[\! \begin{array}{c} \vdots \\ y_{n,m-2} \\ y_{n,m} \\ y_{n,m+2} \\ y_{n,m+4}\\ \vdots \end{array}\!\right]}_{\underline{y}(n)} \!=\! \underbrace{\left[\! \begin{array}{ccccccccccc} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ \cdots & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \cdots\\ & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{array}\!\right]}_{\mathbf{C}_K} \underbrace{\left[\! \begin{array}{c} \vdots \\ y_{n,m-2} \\ y_{n-1,m-1} \\ y_{n,m} \\ y_{n-1,m+1} \\ y_{n,m+2} \\ y_{n-1,m+3} \\ y_{n,m+4}\\ \vdots \end{array}\!\right]}_{\underline{x}_K(n)}$$

The state transition matrix $\mathbf{A}_K$ may be obtained by first performing a one-step time update,

$$y_{n+2,m} = y_{n+1,m-1}+y_{n+1,m+1}-y_{n,m}+\underline{\beta}_m^T \underline{\upsilon}(n+2),$$

and then expanding the two $n+1$ terms in terms of the state at time $n$ :

$$y_{n+1,m-1} = y_{n,m-2}+y_{n,m}-y_{n-1,m-1}+\underline{\beta}_{m-1}^T\underline{\upsilon}(n+1)$$

$$y_{n+1,m+1} = y_{n,m}+y_{n,m+2}-y_{n-1,m+1}+\underline{\beta}_{m+1}^T\underline{\upsilon}(n+1)$$

$$\protect$$ (E.25)

The intra-grid state update for even $m$ is then given by

$${y_{n+2,m}}$$

$$= y_{n,m-2} - y_{n-1,m-1} + y_{n,m} - y_{n-1,m+1} + y_{n,m+2}$$

$$\quad +\; \underline{\beta}_m^T \underline{\upsilon}(n+2) + (\underline{\beta}_{m-1}+\underline{\beta}_{m+1})^T\underline{\upsilon}(n+1)$$

$$= \begin{array}{c} \left[1, -1, 1, -1, 1\right]\\ \vspace{0.5in} \end{array}\left[\! \begin{array}{l} y_{n,m-2}\\ y_{n-1,m-1}\\ y_{n,m}\\ y_{n-1,m+1}\\ y_{n,m+2}\\ y_{n-1,m+3}\\ \end{array}\!\right]$$

$$+ \left[\underline{\beta}_m^T \quad (\underline{\beta}_{m-1}+\underline{\beta}_{m+1})^T \right] \left[\! \begin{array}{l} \underline{\upsilon}(n+2)\\ \underline{\upsilon}(n+1) \end{array}\!\right]. \protect$$ (E.26)

For odd $m$ , the update in Eq.(E.25) is used. Thus, every other row of $\mathbf{A}_K$ , for time $n+2$ , consists of the vector $[1,-1,1,-1,1]$ preceded and followed by zeros. Successive rows for time $n+2$ are shifted right two places. The rows for time $n+1$ consist of the vector $[1,-1,1]$ aligned similarly:

$$\underbrace{\left[\! \begin{array}{l} \qquad\vdots\\ y_{n+1,m-1}\\ y_{n+2,m}\\ y_{n+1,m+1}\\ y_{n+2,m+2}\\ \qquad\vdots \end{array}\!\right]}_{\underline{x}_K(n+2)} \!\leftarrow\! \underbrace{\left[\! \begin{array}{rrrrrrrrrrr} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \cdots & 1 & -1 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 1 & -1 & 1 & -1 & 1 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & -1 & 1 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & -1 & 1 & -1 & 1 & \cdots\\ & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{array}\!\right]}_{\mathbf{A}_K} \! \underbrace{\left[\! \begin{array}{l} \qquad\vdots\\ y_{n,m-2}\\ y_{n-1,m-1}\\ y_{n,m}\\ y_{n-1,m+1}\\ y_{n,m+2}\\ y_{n-1,m+3}\\ \qquad\vdots \end{array}\!\right]}_{\underline{x}_K(n)}$$

From Eq.(E.26) we also see that the input matrix $\mathbf{B}_K$ is given as defined in the following expression:

$$\underbrace{\left[\! \begin{array}{l} \qquad\vdots\\ y_{n+1,m-1}\\ y_{n+2,m}\\ y_{n+1,m+1}\\ y_{n+2,m+2}\\ y_{n+1,m+3}\\ y_{n+2,m+4}\\ y_{n+1,m+5}\\ \qquad\vdots \end{array}\!\right]}_{\underline{x}_K(n+2)} \leftarrow \underbrace{\left[\! \begin{array}{cc} \vdots & \vdots \\ 0 & \underline{\beta}_{m-1}^T \\ [5pt] \underline{\beta}_m^T & \quad \underline{\beta}_{m-1}^T + \underline{\beta}_{m+1}^T \\ [5pt] 0 & \underline{\beta}_{m+1}^T \\ [5pt] \underline{\beta}_{m+2}^T & \underline{\beta}_{m+1}^T + \underline{\beta}_{m+3}^T \\ [5pt] 0 & \underline{\beta}_{m+3}^T \\ [5pt] \underline{\beta}_{m+4}^T & \underline{\beta}_{m+3}^T + \underline{\beta}_{m+5}^T \\ [5pt] 0 & \underline{\beta}_{m+5}^T \\ [5pt] \vdots & \vdots \end{array}\!\right]}_{\mathbf{B}_K} \underbrace{\left[\! \begin{array}{c} \underline{\upsilon}(n+2)\\ \underline{\upsilon}(n+1) \end{array}\!\right]}_{\underline{u}(n+2)}.$$