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3 Optical Signal Attenuation and Dispersion
optical fiber that has an attenuation of 0.4 dB/km at 1310 nm. Suppose an engineer
wants to find the optical output power P out if 200 nW of optical power is launched
into the fiber. First express the input power in dBm units:
P in (dBm) = 10 log
P in (W)
1 mW
= 10 log
200 × 10
−6 W
1 × 10 −3 W
= −7.0 dBm
From Eq. (3.3) with P(0) = P in and P(z) = P out the output power level (in dBm)
at z = 30 km is
P out (dBm) = 10 log
P out (W)
1 mW
= 10 log
P in (W)
1 mW
− αz
= −7.0 dBm − (0.4 dB/km)(30 km) = −19.0 dBm
In unit of watts, the output power is
P(30 km) = 10
−19.0/10
(1 mW) = 12.6 × 10
−3 mW = 12.6 μW.
Drill Problem 3.1 A 50 km long optical fiber has an attenuation of 0.25 dB/km
at 1550 nm. If 100 μW of optical power is launched into the fiber, show that
the power emerging at the fiber output is −32.5 dBm or 0.56 μW.
Drill Problem 3.2 An optical fiber loses 75% of the optical power traversing
the fiber after 25 km. Using the left-hand side of Eq. (3.3) with z = 25 km and
P(z) = 0.25 P(0), show that the attenuation is α = 0.25 dB/km.
3.1.2 Absorption of Optical Power
Absorption of optical power is caused by three different mechanisms:
1. Absorption by atomic defects in the glass composition.
2. Extrinsic absorption by impurity atoms in the glass material.
3. Intrinsic absorption by the basic constituent atoms of the fiber material.
Atomic defects are imperfections in the atomic structure of the fiber material.
Examples of these defects include missing molecules, high-density clusters of
atom groups, or oxygen defects in the glass structure. Usually, absorption losses
3 Optical Signal Attenuation and Dispersion
optical fiber that has an attenuation of 0.4 dB/km at 1310 nm. Suppose an engineer
wants to find the optical output power P out if 200 nW of optical power is launched
into the fiber. First express the input power in dBm units:
P in (dBm) = 10 log
P in (W)
1 mW
= 10 log
200 × 10
−6 W
1 × 10 −3 W
= −7.0 dBm
From Eq. (3.3) with P(0) = P in and P(z) = P out the output power level (in dBm)
at z = 30 km is
P out (dBm) = 10 log
P out (W)
1 mW
= 10 log
P in (W)
1 mW
− αz
= −7.0 dBm − (0.4 dB/km)(30 km) = −19.0 dBm
In unit of watts, the output power is
P(30 km) = 10
−19.0/10
(1 mW) = 12.6 × 10
−3 mW = 12.6 μW.
Drill Problem 3.1 A 50 km long optical fiber has an attenuation of 0.25 dB/km
at 1550 nm. If 100 μW of optical power is launched into the fiber, show that
the power emerging at the fiber output is −32.5 dBm or 0.56 μW.
Drill Problem 3.2 An optical fiber loses 75% of the optical power traversing
the fiber after 25 km. Using the left-hand side of Eq. (3.3) with z = 25 km and
P(z) = 0.25 P(0), show that the attenuation is α = 0.25 dB/km.
3.1.2 Absorption of Optical Power
Absorption of optical power is caused by three different mechanisms:
1. Absorption by atomic defects in the glass composition.
2. Extrinsic absorption by impurity atoms in the glass material.
3. Intrinsic absorption by the basic constituent atoms of the fiber material.
Atomic defects are imperfections in the atomic structure of the fiber material.
Examples of these defects include missing molecules, high-density clusters of
atom groups, or oxygen defects in the glass structure. Usually, absorption losses