6.5.3 Comparison with the Waveguide Method

For that we reconsider (6.32). By comprising the four lateral
vectors
** e_{x}(z)**,

(6.59) |

(6.60) |

(6.61) |

(6.62) |

(6.63) |

(6.64) |

(6.65) |

The solution (6.94) of the ODE
system (6.93) corresponds to the solution of the
*eigenvalue problem* occurring in the waveguide
model [187, eq. (33)]. Since the matrix exponential can also be
evaluated by an eigenvalue decomposition,^{i} both
methods--waveguide and zeroth-order differential--involve the same numerical
operations and thus require the same computational costs. All other steps
like matching the solutions of adjacent layers to determine
the coefficient vectors
** c_{l}^{}** of (6.94) and
thereby propagating the
solution through the whole simulation interval are equal. This proves
theoretically that the differential method comprises the
waveguide model as a sub-class, i.e., as the zeroth-order approximation.

As a direction for future work it would be interesting to compare both methods
numerically. For a fair and correct benchmark a zeroth-order discretization
scheme should be implemented for the differential method. This can be done
most efficiently by determining the solution (6.94)
of (6.93) since otherwise the higher order discretization
is compared with the zeroth-order discretization of the waveguide model.
Our implementation is not optimized for that purpose and the benchmark could
potentially yield misleading results.

- ... obtain
^{g} - Also for vertically
dependent matrices the vector
can be eliminated from (6.89) yielding Future research has to be done to investigate whether a direct solution of the above second-order ODE is advantageous to our implementation which solves the corresponding first-order ODE.**h**(*z*) - ....
^{h} - The eigenvalue
decomposition of a normal matrix
is given by
**=**, whereby is a unitary matrix comprising the eigenvectorsand is a diagonal matrix with the eigenvalues as diagonal entries. The square root is defined as whereby the unitary property**u**_{j}=**u**_{j}**=**of has been used and the square root of the diagonal matrix is simply given by taking the square root of the diagonal entries . Numerically more efficient and stable methods exist to calculate the square root [202, pp. 572-577]. Note that any matrix polynomial can be calculated similarly to the square root, and thus a broad class of functions--including transcendental ones like the exponential--can be calculated from their Taylor series expansions. - ... decomposition,
^{i} - See footnote f on previous page.

1998-04-17