282
7 Optical Receiver Operation
From Fig. 7.9 it follows that Q = 6 (an exact evaluation yields Q = 5.99781),
which gives a signal-to-noise ratio of 12, or 10.8 dB [i.e., 10 1og(S/N) = 10 log 12
= 10.8 dB].
Now consider the special case when σ off = σ on = σ and b off = 0, so that b on =
V. Then, from Eq. (7.14) it follows that the threshold voltage υ th = V /2, so that
Q = V /2σ. Because σ is usually called the rms noise, the ratio V /σ is the peak
signal-to-rms-noise ratio. In this case, Eq. (7.13) becomes
P e (σ on = σ of f ) =
1
2
1 − erf
V
2
√
2 σ
(7.16)
Equation 7.16 demonstrates the exponential behavior of the probability of error as
a function of the signal-to-noise ratio. By increasing V /σ by 2, that is, doubling SNR
(a 3-dB power increase), the BER decreases by 10
4 . Thus, there exists a narrow range
of signal-to-noise ratios above which the error rate is tolerable and below which a
highly unacceptable number of errors occur. The SNR at which this transition occurs
is called the threshold level. In general, a performance safety margin of 3 dB is
included in the transmission link design to ensure that this BER threshold is not
exceeded when system parameters such as transmitter output, line attenuation, or
noise floor vary with time.
Example 7.6 Figure 7.10 shows a plot of the BER expression from Eq. (7.16) as a
function of the signal-to- noise ratio. How many bits are misinterpreted in a standard
DS1 telephone rate of 1.544 Mb/s for SNR values of 8.5 and 12.0?
Solution If the signal-to-noise ratio is 8.5 (18.6 dB) then P e = 10
−5 . If this is the
received signal level for a standard DS1 telephone rate of 1.544 Mb/s, this BER results
in a misinterpreted bit every 0.065 s, which is highly unsatisfactory. However, by
increasing the signal strength so that V/σ = 12.0 (21.6 dB), the BER decreases to
P e = 10
−9 . For the DS1 case, this means that a bit is misinterpreted every 650 s (or
11 min), which, in general, is tolerable.
Drill Problem 7.3 Figure 7.10 shows a plot of the BER expression from
Eq. (7.16) as a function of the signal-to-noise ratio. Verify that for a highspeed network that operates at 622 Mb/s, a required BER of 10
−12 means a
value of V/σ = 22.3 dB is necessary.
7.2.2 Specifying Receiver Sensitivity
Optical communication systems use a BER value to specify the performance requirements for a particular transmission link application. For example, SONET/SDH
7 Optical Receiver Operation
From Fig. 7.9 it follows that Q = 6 (an exact evaluation yields Q = 5.99781),
which gives a signal-to-noise ratio of 12, or 10.8 dB [i.e., 10 1og(S/N) = 10 log 12
= 10.8 dB].
Now consider the special case when σ off = σ on = σ and b off = 0, so that b on =
V. Then, from Eq. (7.14) it follows that the threshold voltage υ th = V /2, so that
Q = V /2σ. Because σ is usually called the rms noise, the ratio V /σ is the peak
signal-to-rms-noise ratio. In this case, Eq. (7.13) becomes
P e (σ on = σ of f ) =
1
2
1 − erf
V
2
√
2 σ
(7.16)
Equation 7.16 demonstrates the exponential behavior of the probability of error as
a function of the signal-to-noise ratio. By increasing V /σ by 2, that is, doubling SNR
(a 3-dB power increase), the BER decreases by 10
4 . Thus, there exists a narrow range
of signal-to-noise ratios above which the error rate is tolerable and below which a
highly unacceptable number of errors occur. The SNR at which this transition occurs
is called the threshold level. In general, a performance safety margin of 3 dB is
included in the transmission link design to ensure that this BER threshold is not
exceeded when system parameters such as transmitter output, line attenuation, or
noise floor vary with time.
Example 7.6 Figure 7.10 shows a plot of the BER expression from Eq. (7.16) as a
function of the signal-to- noise ratio. How many bits are misinterpreted in a standard
DS1 telephone rate of 1.544 Mb/s for SNR values of 8.5 and 12.0?
Solution If the signal-to-noise ratio is 8.5 (18.6 dB) then P e = 10
−5 . If this is the
received signal level for a standard DS1 telephone rate of 1.544 Mb/s, this BER results
in a misinterpreted bit every 0.065 s, which is highly unsatisfactory. However, by
increasing the signal strength so that V/σ = 12.0 (21.6 dB), the BER decreases to
P e = 10
−9 . For the DS1 case, this means that a bit is misinterpreted every 650 s (or
11 min), which, in general, is tolerable.
Drill Problem 7.3 Figure 7.10 shows a plot of the BER expression from
Eq. (7.16) as a function of the signal-to-noise ratio. Verify that for a highspeed network that operates at 622 Mb/s, a required BER of 10
−12 means a
value of V/σ = 22.3 dB is necessary.
7.2.2 Specifying Receiver Sensitivity
Optical communication systems use a BER value to specify the performance requirements for a particular transmission link application. For example, SONET/SDH