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