Page 91 - Fiber Optic Communications Fund

72 Fiber Optic Communications

Intrinsic Absorption
This loss is caused by the interaction of light with pure silica. An electron in the silica molecule absorbs light
and it makes a transition from one electron state to another. This kind of resonance occurs in the ultraviolet
region (< 0.4 μm) for silica and the tail of the absorption band extends throughout the visible spectrum.
A photon could interact with a molecule, causing a change in its vibrational state. This also leads to photon
absorption or optical power loss. This kind of vibrational resonance occurs in the far infrared region > 7 μm,
and the tail of the vibrational resonances can be seen in Fig. 2.34 for > 1.6 μm.

Extrinsic Absorption
This loss is caused by the interaction of light with impurities in silica. Metal impurities such as Cu, Fe,
Cr, Ni and V lead to a strong signal attenuation. These impurities can be reduced to less than one part in
10 10 by glass-refining techniques such as vapor-phase oxidation [23]. One of the major sources of extrinsic
absorption is the water vapors present in silica fibers. The OH ion of the water vapor is bonded into the
glass structure and has a fundamental vibrational resonance at 2.73 μm. Its overtones and combination tones
with the fundamental silica vibrational resonances lead to strong absorption at 1.38, 1.24, 0.95, and 0.88 μm
wavelengths. As shown in Fig. 2.34, the absorption at 1.31 μm is the strongest and its tail at 1.3 μmwas the
main hurdle for the development of fiber-optic communication systems at 1.3 microns. Efforts have been
made to reduce the absorption at 1.31 μm to less than about 0.35 dB/km by reducing the water content in the
glass [24, 25]. The majority of fiber-optic systems operate around the wavelength windows centered at 1.3 μm
and 1.55 μm. This is because the window centered at 1.3 μm has the lowest dispersion for a standard SMF
and the window at 1.55 μm has the lowest loss.

Example 2.8
−1
A fiber of length 80 km has a loss coefficient of 0.046 km . Find the total loss. If the power launched to this
fiber is 3 dBm, find the output power in mW and dBm units.

Solution:
The loss per unit length (dB/km) = 4.343 × 0.046 = 0.2 dB/km. Total loss = 0.2 × 80 = 16 dB. From
Eq. (2.124), we have
P(mW)
P(dBm)= 10 log . (2.179)
10
1mW
From Eq. (2.175), we find
P (mW)= P (mW) exp (−L). (2.180)
out in
Dividing Eq. (2.180) by 1 mW and taking logarithms, we find
{ }
P (mW) exp (−L)
in
P (dBm)= 10 log 10
out
1mW
= P (dBm)+ 10 log exp (−L)
10
in
P out
= P (dBm)+ 10 log . (2.181)
in 10
P
in