Page 317 - Fiber Optic Communications Fund

298 Fiber Optic Communications

Example 6.15
For a Fabry–Perot amplifier, show that the round-trip amplitude gain is
√ √
G max − G min
s
RG = √ √ ,
G + G
max min
where G and G are the maximum and minimum values of G.
max min
Solution:
From Eq. (6.135), we have
√
G 1 + RG
max s
√ = = x,
G min 1 − RG s
or

(1 − RG )x =(1 + RG ),
s s
x − 1
RG =
s
x + 1
√ √
G max − G min
= √ √ .
G + G
max min

Exercises

6.1 Explain the meaning of (a) signal–ASE beating noise and (b) ASE–ASE beating noise.
6.2 An optical amplifier operating at 1300 nm has a mean noise power per unit frequency interval per
polarization (single-sided) of −125 dBm/Hz. Calculate the noise figure. Assume G = 30 dB.
(Ans: 6 dB.)

6.3 The output of an amplifier passes through an ideal optical band-pass filter of bandwidth (f ) 20 GHz, a
o
photodetector of responsivity 0.9 A/W, and an ideal electrical low-pass filter of bandwidth (f ) 8.5 GHz.
e
The amplifier input power is −23 dBm and its gain is 20 dB. The OSNR at the amplifier output (in a
bandwidth of 12.49 GHz) is 15 dB. A polarizer is placed just before the photodetector, which blocks
one of the polarization modes. Find (a) the variance of the signal–ASE beat noise current and (b) the
variance of the ASE–ASE beat noise current. Repeat (a) and (b) if the polarizer is absent.
2
2
2
(Ans: (a) 8.75 × 10 −9 A ;(b) 8.73 × 10 −11 A . When the polarizer is absent, (a) 8.75 × 10 −9 A ;
2
(b) 1.746 × 10 −10 A .)
6.4 State the difference between OSNR and electrical SNR.