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8 Digital Optical Fiber Links
enclosed in underground ducts or in buildings. However it can increase periodically
to over 1 ps/
√
km for outside cables that are suspended on poles, because such cables
are subject to wide variations in temperature, wind-induced stresses, and elongations
caused by ice loading.
To have a power penalty of less than 1.0 dB, the pulse spreading τ PMD resulting
from polarization mode dispersion must on the average be less than 10% of a bit
period T b . Using Eq. (3.40) this condition is given by
τ P M D = D P M D
√
L < 0.1T b
(8.20)
Example 8.7 Consider a 100-km long fiber for which D PMD = 0.5 ps/
√
km. What
is the maximum possible data rate for an NRZ-encoded signal if the pulse spread can
be no more than 10% of a pulse width?
Solution From Eq. (8.20) the pulse spread over the 100-km distance is τ PMD =
5.0 ps. Because this pulse spread can be no more than 10% of a pulse width, it follows
that
τ PMD = 5.0 ps ≤ 0.1 T b
Therefore the maximum NRZ bit rate is 1/T b = 0.1/(5 ps) = 20 Gb/s.
8.2.3 Extinction Ratio Power Penalties
The extinction ratio r e in a laser is defined as the ratio of the optical power level P 1
for a logic 1 to the power level P 0 for a logic 0, that is, r e = P 1 /P 0 . Ideally one would
like the extinction ratio to be infinite, so that there would be no power penalty from
this condition. In this case, if P ave is the average power, then P 0 = 0 and P 1 = 2P ave
= P ideal . However, the extinction ratio must be finite in an actual system in order to
reduce the rise time of laser pulses. That is, the laser must be slightly on during a
zero pulse.
Letting P 1−ER and P 0−ER be the 1 and 0 power levels, respectively, with a finite
extinction ratio, and defining r e = P 1−ER /P 0−ER , the average power is
P ave =
P 1−E R + P 0−E R
2
= P 0−E R
r e + 1
2
= P 1−E R
r e + 1
2r e
(8.21)
When receiver thermal noise dominates, then the 1 and 0 noise powers are equal
and independent of the signal level. In this case, letting P 0 = 0 and P 1 = 2P ave , the
power penalty given by Eq. (8.18) becomes
P P E R = −10 log
P 1−E R − P 0−E R
P 1
= −10 log
r e − 1
r e + 1
(8.22)
8 Digital Optical Fiber Links
enclosed in underground ducts or in buildings. However it can increase periodically
to over 1 ps/
√
km for outside cables that are suspended on poles, because such cables
are subject to wide variations in temperature, wind-induced stresses, and elongations
caused by ice loading.
To have a power penalty of less than 1.0 dB, the pulse spreading τ PMD resulting
from polarization mode dispersion must on the average be less than 10% of a bit
period T b . Using Eq. (3.40) this condition is given by
τ P M D = D P M D
√
L < 0.1T b
(8.20)
Example 8.7 Consider a 100-km long fiber for which D PMD = 0.5 ps/
√
km. What
is the maximum possible data rate for an NRZ-encoded signal if the pulse spread can
be no more than 10% of a pulse width?
Solution From Eq. (8.20) the pulse spread over the 100-km distance is τ PMD =
5.0 ps. Because this pulse spread can be no more than 10% of a pulse width, it follows
that
τ PMD = 5.0 ps ≤ 0.1 T b
Therefore the maximum NRZ bit rate is 1/T b = 0.1/(5 ps) = 20 Gb/s.
8.2.3 Extinction Ratio Power Penalties
The extinction ratio r e in a laser is defined as the ratio of the optical power level P 1
for a logic 1 to the power level P 0 for a logic 0, that is, r e = P 1 /P 0 . Ideally one would
like the extinction ratio to be infinite, so that there would be no power penalty from
this condition. In this case, if P ave is the average power, then P 0 = 0 and P 1 = 2P ave
= P ideal . However, the extinction ratio must be finite in an actual system in order to
reduce the rise time of laser pulses. That is, the laser must be slightly on during a
zero pulse.
Letting P 1−ER and P 0−ER be the 1 and 0 power levels, respectively, with a finite
extinction ratio, and defining r e = P 1−ER /P 0−ER , the average power is
P ave =
P 1−E R + P 0−E R
2
= P 0−E R
r e + 1
2
= P 1−E R
r e + 1
2r e
(8.21)
When receiver thermal noise dominates, then the 1 and 0 noise powers are equal
and independent of the signal level. In this case, letting P 0 = 0 and P 1 = 2P ave , the
power penalty given by Eq. (8.18) becomes
P P E R = −10 log
P 1−E R − P 0−E R
P 1
= −10 log
r e − 1
r e + 1
(8.22)