Something I read a while ago, about constructing an approximation to a
conservative system by performing the difference equation update in a
\emph{ping--pong} fashion, i.e. using speed to update position and
then the new position to update speed, instead of performing the
updates in parallel. Supposedly, this preserves the \emph{symplectic
structure}. I lost the reference, so this is an attempt to
reconstruct it.
This might be related with approximations to 2nd order CT filters
mentioned in Stilson's PhD thesis[1], where two integrators are
replaced with a backward and forward difference each. A similar thing
happens here: the update equations are no longer parallel, since
$x[k+1]$ (first integrator, BP output) is used in the computation of
$y[k+1]$ (second integrator, LP output).
It's probably best to first define what \emph{symplectic structure}
means, and then to see how it can be preserved in the analog to
digital conversion. What I remember is that it is important to be
able to factor the update into two successive triangular updates.
This is exactly what happens in the mixed FW/BW SVF discretization.
At full resonance $k=0$ (we're talking about lossless systems only) in
the update
$$ \begin{array}{ccl}
z x_1 &=& x_1 - a (k x_1 + x_2 - i) \\
z x_2 &=& x_2 + a z x_1 \\ \end{array},$$
we have two consecutive
operations which hide behind the presence of $z x_1$ in the equation
(as opposed to $x_1)$. This succession of updates can be represented
by two triangular matrices.
$$
\begin{array}{ccc} A = L U, & L = \matrix{1 & 0 \\ a & 1}, & U =
\matrix{1 &-a \\ 0 & 1} \end{array}.$$
Note that this matrix is not orthogonal or $A^TA \neq I$. However, it
is symplectic. A 2D discrete system with feedback matrix $A$ is
symplectic if $A^T J A = J$ with $$J=\matrix{0 & 1 \\ -1 & 0}.$$ This
property is satisfied by the conservative forward/backward SVF. Note
that $J^2 = -I$.
Because $\det A = 1$ and $A \in R^{2 \times 2}$, if the poles are
complex conjugate they have to be magnitude $1$ so the system is still
stable.
So, is a symplectic system always stable? In general, a $A \in R^{2N
\times 2N}$ is symplectic if $A^T J_N A = J$ where $$J_N = \matrix{0 &
I \\ -I_N & 0}$$ where $I$ is the $N \times N$ identity matrix. From
this we have $\det A = 1$. So it is conservative in that state energy
is preserved, but that doesn't mean the individual 2D subsystems
should be energy--preserving. In [2], p282 it is explained how for
each eigenvalue $\lambda$, the inverse $\lambda^{-1}$ is also an
eigenvalue.
An example of a 4D symplectic matrix can be constructed from a 2D
symplectic matrix $A_1^T J_1 A_1 = J_1$ by constructing the matrix
$$A'_2 = \matrix{ r A_1 & 0 \\ 0 & r^{-1} A_1 }.$$ Straightforward
computation shows that $A'_2^T J'_2 A'_2 = J'_2$ where $$J'_2 =
\matrix{0 & J_1 \\ -J_1 & 0},$$ which is a permutation of $J_2$, meaning
$J'_2 = P^T J_2 P$. From this follows that $A_2 = P^T A'_2 P$ is
symplectic. Since $P$ is orthogonal, $A_2$ and $A'_2$ have the same
eigenvalues. The eigenvalues of $A_1$ are unit norm complex conjucate
which makes the eigenvalues of $A_2$ complex conjugate pairs of
magnitude $r$ and $r^{-1}$ respectively.
% [1] https://ccrma.stanford.edu/~stilti/papers/TimStilsonPhDThesis2006.pdf
% [2] http://mitpress.mit.edu/sites/default/files/titles/content/sicm/book.html