More General Velocity Excitations

From Eq.(E.11), it is clear that initializing any single K variable $y_{n,m}$ corresponds to the initialization of an infinite number of W variables $y^{+}_{n,m}$ and $y^{-}_{n,m}$ . That is, a single K variable $y_{n,m}$ corresponds to only a single column of $\mathbf{T}^{-1}$ for only one of the interleaved grids. For example, referring to Eq.(E.11), initializing the K variable $y_{n-1,m}$ to -1 at time $n$ (with all other $y_{n,m}$ intialized to 0) corresponds to the W-variable initialization

$$\begin{eqnarray*} y^{+}_{n,m-(2\mu+1)}&=&+1, \quad \mu =0,1,2,\cdots\\ y^{-}_{n,m-(2\mu+1)}&=&-1, \quad \mu =0,1,2,\cdots \end{eqnarray*}$$

with all other W variables being initialized to zero. In view of earlier remarks, this corresponds to an impulsive velocity excitation on only one of the two subgrids. A schematic depiction from $\mu = m-5$ to $m+5$ of the W variables at time $n$ is as follows:

$$\begin{array}{crrrrr\vert rrrrrrc} \cdots & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & -1 & 0 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ & & & & & & m & & & & & \mu & \rightarrow \end{array}$$ (E.14)

Below the solid line is the sum of the left- and right-going traveling-wave components, i.e., the corresponding K variables at time $n$ . The vertical lines divide positions $\mu=m-1$ and $\mu=m$ . The initial displacement is zero everywhere at time $n$ , consistent with an initial velocity excitation. At times $\nu=n+1,n+2,n+3,n+4$ , the configuration evolves as follows:

$$\begin{array}{crrrrr\vert rrrrrrc} \cdots & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \end{array}$$ (E.15)

$$\begin{array}{crrrrr\vert rrrrrrc} \cdots & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots \end{array}$$ (E.16)

$$\begin{array}{crrrrr\vert rrrrrrc} \cdots & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & \cdots \end{array}$$ (E.17)

$$\begin{array}{crrrrr\vert rrrrrrc} \cdots & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & \cdots\\ \cdots & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & \cdots \end{array}$$ (E.18)

The sequence $[\dots,1,0,1,0,1,\dots]$ consists of a dc (zero-frequency) component with amplitude $1/2$ , plus a sampled sinusoid of amplitude $1/2$ oscillating at half the sampling rate $f_s=1/T$ . The dc component is physically correct for an initial velocity point-excitation (a spreading square pulse on the string is expected). However, the component at $f_s/2$ is usually regarded as an artifact of the finite difference scheme. From the DW interpretation of the FDTD scheme, which is an exact, bandlimited physical interpretation, we see that physically the component at $f_s/2$ comes about from initializing velocity on only one of the two interleaved subgrids of the FDTD scheme. Any asymmetry in the excitation of the two interleaved grids will result in excitation of this frequency component.

Due to the independent interleaved subgrids in the FDTD algorithm, it is nearly always non-physical to excite only one of them, as the above example makes clear. It is analogous to illuminating only every other pixel in a digital image. However, joint excitation of both grids may be accomplished either by exciting adjacent spatial samples at the same time, or the same spatial sample at successive times instants.

In addition to the W components being non-local, they can demand a larger dynamic range than the K variables. For example, if the entire semi-infinite string for $m<0$ is initialized with velocity $2X/T$ , the initial displacement traveling-wave components look as follows:

$$\begin{array}{crrrrrr\vert rrrrrc} \cdots & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & -6 & -5 & -4 & -3 & -2 & -1 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \end{array}$$ (E.19)

and the variables evolve forward in time as follows:

$$\begin{array}{crrrrrr\vert rrrrrc} \cdots & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & -5 & -4 & -3 & -2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 2 & 2 & 2 & 2 & 2 & 2 & 1 & 0 & 0 & 0 & 0 & \cdots \end{array}$$ (E.20)

$$\begin{array}{crrrrrr\vert rrrrrc} \cdots & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 & 0 & \cdots\\ \cdots & -4 & -3 & -2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 4 & 4 & 4 & 4 & 4 & 3 & 2 & 1 & 0 & 0 & 0 & \cdots \end{array}$$ (E.21)

$$\begin{array}{crrrrrr\vert rrrrrc} \cdots & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 & \cdots\\ \cdots & -3 & -2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \hline\\ [-1em] \cdots & 6 & 6 & 6 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 & \cdots \end{array}$$ (E.22)

Thus, the left semi-infinite string moves upward at a constant velocity of 2, while a ramp spreads out to the left and right of position $\mu=m$ at speed $c$ , as expected physically. By Eq.(E.9), the corresponding initial FDTD state for this case is

$$\begin{eqnarray*} y_{n,\mu} &=&0, \quad \mu\in\mathbb{Z}\\ y_{n-1,m-1} &=& -1,\\ y_{n-1,\mu} &=& -2, \quad \mu<m-1, \end{eqnarray*}$$

where $\mathbb{Z}$ denotes the set of all integers. While the FDTD excitation is also not local, of course, it is bounded for all $\mu$ .

Since the traveling-wave components of initial velocity excitations are generally non-local in a displacement-based simulation, as illustrated in the preceding examples, it is often preferable to use velocity waves (or force waves) in the first place [450].

Another reason to prefer force or velocity waves is that displacement inputs are inherently impulsive. To see why this is so, consider that any physically correct driving input must effectively exert some finite force on the string, and this force is free to change arbitrarily over time. The ``equivalent circuit'' of the infinitely long string at the driving point is a ``dashpot'' having real, positive resistance $2R=2\sqrt{K\epsilon }$ . The applied force $f(t)$ can be divided by $2R$ to obtain the velocity $v(t)$ of the string driving point, and this velocity is free to vary arbitrarily over time, proportional to the applied force. However, this velocity must be time-integrated to obtain a displacement $y(t)$ . Therefore, there can be no instantaneous displacement response to a finite driving force. In other words, any instantaneous effect of an input driving signal on an output displacement sample is non-physical except in the case of a massless system. Infinite force is required to move the string instantaneously. In sampled displacement simulations, we must interpret displacement changes as resulting from time-integration over a sampling period. As the sampling rate increases, any physically meaningful displacement driving signal must converge to zero.