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State Transformations
In previous work, time-domain adaptors (digital filters) converting
between K variables and W variables have been devised
[224]. In this section, an alternative approach is
proposed. Mapping Eq.(E.7) gives us an immediate conversion from W
to K state variables, so all we need now is the inverse map for any
time
. This is complicated by the fact that non-local spatial
dependencies can go indefinitely in one direction along the string, as
we will see below. We will proceed by first writing down the
conversion from W to K variables in matrix form, which is easy to do,
and then invert that matrix. For simplicity, we will consider the
case of an infinitely long string.
To initialize a K variable simulation for starting at time
, we
need initial spatial samples at all positions
for two successive
times
and
. From this state specification, the FDTD scheme
Eq.(E.3) can compute
for all
, and so on for
increasing
. In the DW model, all state variables are defined as
belonging to the same time
, as shown in Fig.E.2.
Figure E.2:
DW flow diagram.
|
From Eq.(E.6), and referring to the notation defined in
Fig.E.2, we may write the conversion from W to K variables
as
where the last equality follows from the traveling-wave behavior
(see Fig.E.2).
Figure E.3:
Stencil of the FDTD scheme.
|
Figure E.3 shows the so-called ``stencil'' of the FDTD scheme.
The larger circles indicate the state at time
which can be used to
compute the state at time
. The filled and unfilled circles
indicate membership in one of two interleaved grids [55]. To
see why there are two interleaved grids, note that when
is even,
the update for
depends only on odd
from time
and even
from time
. Since the two W components of
are converted to
two W components at time
in Eq.(E.8), we have that the update for
depends only on W components from time
and positions
.
Moving to the next position update, for
, the state used is
independent of that used for
, and the W components used are
from positions
and
. As a result of these observations, we
see that we may write the state-variable transformation separately for
even and odd
, e.g.,
![$\displaystyle \left[\! \begin{array}{c} \vdots \\ y_{n,m-1}\\ y_{n-1,m}\\ y_{n,m+1}\\ y_{n-1,m+2}\\ y_{n,m+3}\\ y_{n-1,m+4}\\ y_{n,m+5}\\ \vdots \\ \end{array} \!\right] \!=\! \left[\! \begin{array}{cccccccccc} \ddots & & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 0 \\ \cdots & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & \cdots\\ 0 & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \!\right] \left[\! \begin{array}{c} \vdots \\ y^{+}_{n,m-1}\\ y^{-}_{n,m-1}\\ y^{+}_{n,m+1}\\ y^{-}_{n,m+1}\\ y^{+}_{n,m+3}\\ y^{-}_{n,m+3}\\ y^{+}_{n,m+5}\\ y^{-}_{n,m+5}\\ \vdots \end{array} \!\right]. \protect$](img4605.png) |
(E.9) |
Denote the linear transformation operator by
and the K and W state vectors
by
and
, respectively. Then Eq.(E.9) can be restated as
 |
(E.10) |
The operator
can be recognized as the Toeplitz operator
associated with the linear, shift-invariant filter
.
While the present context is not a simple convolution since
is
not a simple time series, the inverse of
corresponds to the
Toeplitz operator associated with
Therefore, we may easily write down the inverted transformation:
![$\displaystyle \left[\! \begin{array}{c} \vdots \\ y^{+}_{n,m-1}\\ y^{-}_{n,m-1}\\ y^{+}_{n,m+1}\\ y^{-}_{n,m+1}\\ y^{+}_{n,m+3}\\ y^{-}_{n,m+3}\\ y^{+}_{n,m+5}\\ y^{-}_{n,m+5}\\ \vdots \end{array} \!\right] \!=\! \left[\! \begin{array}{crrrrrrrrc} \ddots & & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \pm1 \\ \cdots & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & \cdots\\ \cdots & 0 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & \cdots\\ \cdots & 0 & 0 & 1 & -1 & 1 & -1 & 1 & -1 & \cdots\\ \cdots & 0 & 0 & 0 & 1 & -1 & 1 & -1 & 1 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 1 & -1 & 1 & -1 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 1 & \cdots\\ \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & \cdots\\ 0 & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \!\right] \!\! \left[\! \begin{array}{c} \vdots \\ y_{n,m-1}\\ y_{n-1,m}\\ y_{n,m+1}\\ y_{n-1,m+2}\\ y_{n,m+3}\\ y_{n-1,m+4}\\ y_{n,m+5}\\ \vdots \\ \end{array} \!\right] \protect$](img4612.png) |
(E.11) |
The case of the finite string is identical to that of the infinite
string when the matrix
is simply ``cropped'' to a finite square size
(leaving an isolated 1 in the lower right corner); in such cases,
as given above is simply cropped to the same size, retaining its
upper triangular
structure. Another interesting set of cases
is obtained by inserting a 1 in the lower-left corner of the cropped
matrix to make it circulant; in these cases, the
matrix
contains
in every position for even
, and
is singular for odd
(when there is one zero eigenvalue).
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