Page 183 - Introduction to Information Optics

168 3. Communication with Optics

Example 3.3. A step index optical fiber has core refractive index n l = 1.5,
cladding refractive index n 2 — 1.485, length L = 1 km. Calculate the maximum
bit rate for this fiber.
Solve: The maximum bit rate, B, for this fiber is

D __ _ _ ..... 3 ^ 20 Mbos
At" «, L/c(n Jn 2 - 1) 1.5 -10 m/3 • 10 m/s(1.5/1.485 - 1) ~ " *""

From this example, one can see that B is much less than the optical carrier
14
frequency; that is, in the order of 10 Hz. To alleviate this problem, we need
to reduce the time difference among different propagating routes. Fortunately,
there is a case in which only one route is allowed to propagate in the fiber (i.e.,
single mode fiber) so that much higher bandwidth can be achieved. However,
the simple geometric optics theory cannot fully explain this phenomenon,
which must be clarified by the more precise wave-optics theory, described in
the next section.

3.2.2. WAVE-OPTICS APPROACH

Since the transversal dimension of optical fiber is comparable with its
wavelength, the more precise wave-optics theory is needed to explain all the
phenomena happening in the fiber; in particular, the mode theory.
The wave-optics approach starts with the well-known Maxwell equation.
Optical fiber is a dielectric material so the free charge density is p = 0 and free
current density j = 0. In addition, assume the light wave is a harmonic wave.
Note that the general case can be treated as the weighted summation based on
the Fourier transform for the linear system. Under these assumptions, the
electric field of the light field, E, satisfies the following wave equation:

2
2
V £ + n ^ = 0, (3.11)
2
2
where kl = co /C is the wave number, co is the angular frequency, C is the light
speed in vacuum, n = v/ iu r£ r is the refractive index of the fiber material — and it
2
may be a function of angular frequency (i.e., n = «[co]), and V is the Laplacian
operator. Due to the cylindrical symmetry of the optical fiber, as shown in Fig.
3.3, it is convenient to solve Eq. (3.11) under cylindrical coordinates. Note that
Eq. (3.11) is a vector differential equation. For simplicity, let us deal with the
z component of the electric field, E z, first. In this case, Eq. (3.11) becomes the
following simplified scalar differential equation:

2
2
V E, + n klE z = 0. (3.12)