14.8 Optical Fiber Network Performance Testing
613
B E R =
N e
B T
(14.16)
Thus the gating time window needed to measure N e = 100 errors is T =
100/(BER × B). For high- speed communications greater than 1 Gb/s, the required
bit-error rate typically needs to be 10
–12 or lower.
Example 14.6 A given 10-Gb/s link is designed to operate with a 10
–12 BER. What
is the gating time window for accumulating 100 bit errors?
Solution From Eq. (14.16) the gating time is T = 100/(BER × B) = 100/(10
–12
× 10
10 ) = 10
4 s ≈ 2.7 h.
However, a 10
–12 BER level may be unacceptable for a 10-Gb/s data rate, so
even lower bit-error rates, such as 10
–15 , may be required to assure customers of
a high grade of service. To accumulate 100 errors for such a BER would require
over 100 days of measurement time. Because this is not practical, modern BER
measuring instruments add an extra precisely calibrated amount of noise into the
system, thereby accelerating the occurrence of errors. The extra noise decreases the
receiver threshold, which increases the probability of errors and thus greatly reduces
the gating time window. Although some accuracy is lost in this method, it reduces
the test times to minutes instead of hours or days.
Example 14.7 In a real communication system, network operators also are interested
in the frame error rate P frame . If the total number of bits in a frame is k and P e is the
BER, then the probability that no bit in the frame has an error is.
1 − P frame = (1 − P e )
k
≈ 1 − P e k
(14.17)
where the approximation results from the condition that P e 1. If P e = 10
–12 , what
is the frame-error rate for an Ethernet frame length of 1518 bytes?
Solution Because there are 8 bits per byte, this frame has 12,144 bits. From
Eq. (14.17) it follows that P frame = P e k = (10
–12 ) × 12,144 = 1.2144 × 10
–8 .
14.8.2 OSNR Measurements
Measuring the SNR and its associated BER is straightforward for single-wavelength
links with no optical amplifiers. However, in multispan optically amplified DWDM
networks, the system performance is limited by the OSNR rather than by the optical
signal power that arrives at the receiver. Although one could demultiplex the incoming
DWDM traffic and then do BER estimates on each individual wavelength channel, an
optical spectrum measurement can be performed with an optical spectrum analyzer
(OSA) to derive the OSNR for each individual channel. The OSNR derived from
613
B E R =
N e
B T
(14.16)
Thus the gating time window needed to measure N e = 100 errors is T =
100/(BER × B). For high- speed communications greater than 1 Gb/s, the required
bit-error rate typically needs to be 10
–12 or lower.
Example 14.6 A given 10-Gb/s link is designed to operate with a 10
–12 BER. What
is the gating time window for accumulating 100 bit errors?
Solution From Eq. (14.16) the gating time is T = 100/(BER × B) = 100/(10
–12
× 10
10 ) = 10
4 s ≈ 2.7 h.
However, a 10
–12 BER level may be unacceptable for a 10-Gb/s data rate, so
even lower bit-error rates, such as 10
–15 , may be required to assure customers of
a high grade of service. To accumulate 100 errors for such a BER would require
over 100 days of measurement time. Because this is not practical, modern BER
measuring instruments add an extra precisely calibrated amount of noise into the
system, thereby accelerating the occurrence of errors. The extra noise decreases the
receiver threshold, which increases the probability of errors and thus greatly reduces
the gating time window. Although some accuracy is lost in this method, it reduces
the test times to minutes instead of hours or days.
Example 14.7 In a real communication system, network operators also are interested
in the frame error rate P frame . If the total number of bits in a frame is k and P e is the
BER, then the probability that no bit in the frame has an error is.
1 − P frame = (1 − P e )
k
≈ 1 − P e k
(14.17)
where the approximation results from the condition that P e 1. If P e = 10
–12 , what
is the frame-error rate for an Ethernet frame length of 1518 bytes?
Solution Because there are 8 bits per byte, this frame has 12,144 bits. From
Eq. (14.17) it follows that P frame = P e k = (10
–12 ) × 12,144 = 1.2144 × 10
–8 .
14.8.2 OSNR Measurements
Measuring the SNR and its associated BER is straightforward for single-wavelength
links with no optical amplifiers. However, in multispan optically amplified DWDM
networks, the system performance is limited by the OSNR rather than by the optical
signal power that arrives at the receiver. Although one could demultiplex the incoming
DWDM traffic and then do BER estimates on each individual wavelength channel, an
optical spectrum measurement can be performed with an optical spectrum analyzer
(OSA) to derive the OSNR for each individual channel. The OSNR derived from