The basis for the application of the previously introduced RTL method is the radial discretization of all cylindrical fibres via a separation into a succession of thin cylindrical layers, each one with its own constant refractive index
n. These layers can be made to extend outside of the cladding in order to take into consideration the effect of the surrounding air (
n = 1). Each thin cylindrical layer could have thickness
δr proportional to each average radius
r which means that given discrete steps as
with
one has
Furthermore, following a cumbersome analysis, it is possible to prove that the system of Equations (6) and (7) can be transformed in a set of four differential Equation (8), relating the equivalent “voltage’ and “current” functions
defined as follows:
where we use the notation
At this point it is noticed that
are continuous functions at the boundaries because the tangential components of electric and magnetic fields
and
on the cylindrical surface are continuous functions passing the boundaries’ of the cylindrical layer. Using the previous relations, the Fourier Transforms of the electromagnetic field components along (
r,
,
β) can be expressed as functions of their equivalent “voltages” and “currents” functions with the auxiliary relations
It becomes evident by inspection that the final Equation (8) represents two coupled electric transmission lines.
2.1. Decoupling the Transmission Line Equations
The prescribed set of Equation (8) constitutes a homogeneous set of ordinary differential equations of
r and considering that the all vectors [
] can be turned into exponential functions of
r given by
, where
are constants, i.e., not functions of
r. Thus, the system (8) can be transformed in an algebraic set of the following four equations
Replacing
,
, we obtain a set of two homogeneous equations
This then leads to the eigenvalue equations
From the standard form of the eigenvalue problem we obtain through the determinant differential equations as follows
Hence the system has two eigenvalues and two mutually excluded or “normal” eigenvectors. The eigenvectors will be found by replacing
by its value. Thus, for
,
and the eigenvector is
While for
,
and the eigenvector becomes
. Their respective “current” eigenvectors are related as
, thus
and
,
. Since the auxiliary
M function has the sign of
l, the set
, for
becomes equal to the set
. Thus, we can consider as a unique solution for the set
and the integer ‘
’varies
, and of course
:
Furthermore,
should be continuous functions at their boundaries although
n(
r) varies from layer to layer. This is achieved via the adjustment
and
, which are continuous functions of
r by definition.
Another option for achieving continuity is to consider the functions
and
. In this case,
and
are also continuous. Thus
Thus, the set of two coupled transmission lines (9) is equivalent to two independent transmission lines (12) and (13).
The two waves represented by the equations of transmission lines (12) and (13), are geometrically normal because the first is related to the magnetic field and the second to the electric field that are geometrically normal for transmitted EM waves. This property is an inherent property of EM modes in optical fibers related to birefringence phenomena. However, the β respective values, for any mode, are always found to be very close and can be considered as practically equal.
2.2. Equivalent Circuits for Cylindrical Layers, Boundary Conditions, and Birefringence
Taking into consideration the transmission line theory, it can be proved that each layer of infinitesimal thickness
δr is equivalent to a T-circuit as the one shown in
Figure 3For
the impedances can be approximated by the equivalent relations
If
, both
are “capacitive” reactances, for
however
becomes “inductive” reactance. For
the approximate respective impedances of the T-circuit are given as
As previously stated, the functions of each layer are continuous at the cylindrical boundaries of the layer, thus if we divide the fiber (including a sufficient number of air layers) in successive thin layers and replace them by their equivalent T-circuits, an overall lossless transmission line is formed with only reactive elements. For given “l”, the “β” values that lead to the resonance of the overall transmission line are the eigenvalues of the whole optical fiber.
When a transmission line is in resonance, at any arbitrary point
of the line, the sum of reactive impedances arising from the successive T-circuits on the left and right sides of
should be equal to zero, thus the equation giving the eigenvalues of the transmission line is the following:
Equation (16) provides the eigenvalues “β” for a given “l”, where are the overall reactive impedances of successive T-circuits on the left and right of , using Equations (14) or (15). The value of is usually given by the core radius. For the same “” the Equations (14) and (15) give usually slightly different values of ‘β’. This phenomenon is called “Birefringence”. For circular step index fibers, the birefringence is negligible; however, for elliptic fibers and fibers of any other non-circular cores, the birefringence phenomenon could be not negligible.
In order to calculate the overall reactive impedances on the left and right of
we should find the impedances for
and for
. As we proceed to 0 or to
, the remaining piece of transmission line becomes “homogeneous”, i.e., its overall reactive impedance is equal to its characteristic impedance given by
Then we must have
For
l =
0 (open circuit at the center of the equivalent transmission line) It is useful to notice that there is an equivalence between our formulation and the classic formulation modes of optical fibers. In particular, for
=
0, the modes (
VM,
IM) are the TM modes, while the modes (
VE,
IE) are the TE modes. For
> 0, the modes (
VM,
IM) are the HE modes, while the modes (
VE,
IE) are their HE birefringence modes. For
<
0 the modes (
VM,
IM) are the EH modes, while the modes (
VE,
IE) are their EH birefringence modes. For any given
, using the resonance technique the
β values of the two birefringence modes can be calculated. The Equation (16) is given as a MATLAB function in
Appendix A.
Let us consider for example a step-index fiber of n1 = 1.54, n2 = 1.47 the VM, VE, fundamental modes for V = 3.3, can be calculated and their β/k0 values are respectively 1.518934962534846 and 1.518340184686295, hence their birefringence is equal to 0.0004947 or 0.0391%. The β/k0 value for the equivalent mode Veq was also calculated and was equal to 1.518638548412019 (that is approximately equal to the mean value of the previous β/k0 values), while the β/k0 value calculated classically by Bessel functions is equal to 1.518642063686336. These β values are very close differing only 0.0002315%.
In the following
Figure 4, the normalized birefringence of the step-index fibers for
n1 = 1.54,
n2 = 1.47, and of
n1 = 1.475,
n2 = 1.47 as functions of V are shown.
We notice that for any V, the normalized birefringence is almost proportional to Δn2 = (n1 − n2)2, thus the birefringence of step-index fibers of very small Δn is negligible. For instance, for a value of V = 2.4, and Δn = 1.54 − 1.47 = 0.07, the birefringence is found to be 0.168 × 0.0049 = 0.0008232 or ~0.055% on the average β, while for Δn = 1.475 − 1.47 = 0.005, the birefringence becomes 0.168 × 0.000025 = 0.000042 or ~0.0028% on the average β. What is remarkable is that our method is sensitive and calculates it.
2.3. Calculating “Voltages” VM, VE and “Currents” IM, IE and Resulting Fields
For any given
, using the resonance technique the
β values of the two birefringence modes can be calculated. These
β values are practically the same, thus we can consider them as equal or we can consider as the proper value of
β the mean value of the two modes. Taking
VM = 1 at the center point of the fiber (
r =
0), the respective value of
IM at the same point can be calculated by the respective terminal impedance. Using the matrix relations between input–output for the equivalent successive T-circuits, the values of
and
at the rest thin cylindrical layers can be calculated. In fact, from the general theory of the telegrapher’s equation we know that the inputs and outputs are associated via a transfer matrix as follows
In Equation (17), the characteristic impedance should be taken as
to fit with the previous analysis. Using the relations
and
the respective values of their birefringence partners can also be calculated for every thin cylindrical layer
ri. Finally, we obtain the actual fields via the relations
A very useful field component for optical fibers is the value of the overall electric field at any thin cylindrical layer of average radius
r that can be calculated by the formula:
After some algebra, this leads to the formula
In the next section, we extend our analysis in certain UOF cases.