11.10 Summary
473
Symbol
Parameter
Value
G
Nominal gain
7.1 dB
P s,out
Nominal output optical power
3.0 dBm
P s,in
Nominal input optical power
−4.1 dBm
11.10 Consider an EDFA with a gain of 26 dB and a maximum output power of 0
dBm.
(a) Compare the output signal levels per channel for 1, 2, 4, and 8 wavelength
channels, where the input power is 1 μW for each signal.
(b) What are the output signal levels per channel in each case if the pump
power is doubled?
Answers to Selected Problems
11.1 (a) R p = 1.26 × 10
27 (electrons/cm
3 )/s (b) g 0 = 7.5 cm
−1
(c) N ph,sat = 1.67 × 10
15 photons/cm
3 (d) N = 1.32 × 10
10 photons/cm
3 .
11.3 P out,sat = 0.693 P amp,sat
11.4 With the 3-dB gain G = 27 dB, then FWHM = 0.50 λ
11.5 (a) PCE (980 nm) ≤ 63.4%; PCE (1475 nm) ≤ 95.5%
11.6 (a) G = 25 dB; (b) P p,in ≥ 785 mW
11.7 σ
2
th = 1.62 × 10
−14 A
2 ; σ
2
shot−s = 2.34 × 10
−14 A
2 ; σ
2
AS E−AS E = 2.26 ×
10
−14 A
2 ; σ
2
s−AS E = 5.47 × 10
−12 A
2 ; σ
2
AS E−AS E = 7.01 × 10
−13 A
2
11.9 Because the slope of the gain-versus -input power curve is –0.5, then for a
6-dB drop in the input signal, the gain increases by +3 dB.
1. Thus at the first amplifier, a −10.1-dBm signal now arrives and experiences a +10.1-dB gain. This gives a 0-dBm output (versus a normal +
3-dBm output).
2. At the second amplifier, the input is now −7.1 dBm (down 3 dB from
the usual −4.1 dBm level). Hence the gain is now 8.6 dB (up 1.5 dB),
yielding an output of
−7.1 dBm + (7.1 + 1.5) dB = 1.5 dBm
3. At the third amplifier, the input is now −5.6 dBm (down 1.5 dB from the
usual −4.1 dBm level). Hence the gain is up 0.75 dB, yielding an output
of
−5.6 dBm + (7.1 + 0.75) dB = 2.25 dBm
4. At the fourth amplifier, the input is now −4.85 dBm (down 0.75 dB from
the usual −4.1 dBm level). Hence the gain is up 0.375 dB, yielding an
output of
473
Symbol
Parameter
Value
G
Nominal gain
7.1 dB
P s,out
Nominal output optical power
3.0 dBm
P s,in
Nominal input optical power
−4.1 dBm
11.10 Consider an EDFA with a gain of 26 dB and a maximum output power of 0
dBm.
(a) Compare the output signal levels per channel for 1, 2, 4, and 8 wavelength
channels, where the input power is 1 μW for each signal.
(b) What are the output signal levels per channel in each case if the pump
power is doubled?
Answers to Selected Problems
11.1 (a) R p = 1.26 × 10
27 (electrons/cm
3 )/s (b) g 0 = 7.5 cm
−1
(c) N ph,sat = 1.67 × 10
15 photons/cm
3 (d) N = 1.32 × 10
10 photons/cm
3 .
11.3 P out,sat = 0.693 P amp,sat
11.4 With the 3-dB gain G = 27 dB, then FWHM = 0.50 λ
11.5 (a) PCE (980 nm) ≤ 63.4%; PCE (1475 nm) ≤ 95.5%
11.6 (a) G = 25 dB; (b) P p,in ≥ 785 mW
11.7 σ
2
th = 1.62 × 10
−14 A
2 ; σ
2
shot−s = 2.34 × 10
−14 A
2 ; σ
2
AS E−AS E = 2.26 ×
10
−14 A
2 ; σ
2
s−AS E = 5.47 × 10
−12 A
2 ; σ
2
AS E−AS E = 7.01 × 10
−13 A
2
11.9 Because the slope of the gain-versus -input power curve is –0.5, then for a
6-dB drop in the input signal, the gain increases by +3 dB.
1. Thus at the first amplifier, a −10.1-dBm signal now arrives and experiences a +10.1-dB gain. This gives a 0-dBm output (versus a normal +
3-dBm output).
2. At the second amplifier, the input is now −7.1 dBm (down 3 dB from
the usual −4.1 dBm level). Hence the gain is now 8.6 dB (up 1.5 dB),
yielding an output of
−7.1 dBm + (7.1 + 1.5) dB = 1.5 dBm
3. At the third amplifier, the input is now −5.6 dBm (down 1.5 dB from the
usual −4.1 dBm level). Hence the gain is up 0.75 dB, yielding an output
of
−5.6 dBm + (7.1 + 0.75) dB = 2.25 dBm
4. At the fourth amplifier, the input is now −4.85 dBm (down 0.75 dB from
the usual −4.1 dBm level). Hence the gain is up 0.375 dB, yielding an
output of