Page 348 - Fiber Optic Communications Fund

Transmission System Design 329

4k Tf e 2
B
2
= 2qI f + + 2R [2P f + (2f − f )f ]
1 1 e R ASE out e ASE o e e
L
4 × 1.38 × 10 −23 × 298 × 7.5 × 10 9
−5
−19
9
= 2 × 1.602 × 10 × 5.32 × 10 × 7.5 × 10 +
200
2
+ 2 × 1.1 × 1.78 × 10 −17 ×[2 × 4.78 × 10 −5 × 7.5 × 10 9
9
9
9
+ 1.78 × 10 −17 ×(2 × 20 × 10 − 7.5 × 10 )× 7.5 × 10 ]
2
= 3.18 × 10 −11 A ,
4k Tf e
B
2
2 2
= 2qI f + + 2R B(2f − f )f
0 0 e ASE o e e
R L
9
= 2 × 1.602 × 10 −19 × 7.83 × 10 −7 × 7.5 × 10 + 4 × 1.38 × 10 −23 × 298 × 7.5 × 10 9
200
2
) ×(2 × 20 × 10 − 7.5 × 10 )× 7.5 × 10
+ 2 × 1.1 ×(1.78 × 10 −17 2 9 9 9
2
= 8.05 × 10 −13 A ,
I − I 0
1
Q =
+
1 0
5.32 × 10 −5 − 7.83 × 10 −7
= √ √
3.18 × 10 −11 + 8.05 × 10 −13
= 8.031.
Example 7.7
A two-stage amplifier with a DCF between the stages needs to be designed. The insertion loss of the DCF
is 7 dB. There are two amplifiers Amp1 and Amp2 with gains G = 8 dB and G = 16 dB, respectively. The
2
1
noise figures of the amplifiers are F n,1 = 7 dB and F n,2 = 5.5 dB. Find the optimum amplifier configuration.
Solution:
Since Amp2 has the lower noise figure, let us first choose Amp2 as the first amplifier. From Eq. (7.122) with
the indices reversed, we have
F n,1
F = F + ,
n,eq n,2
G H
2
F n,1 = 10 F n,1 (dB)∕10 = 5.01,
F = 10 F n,2 (dB)∕10 = 3.54,
n,2
G = 10 G 2 (dB)∕10 = 39.81,
2
H = 10 H(dB)∕10 = 5.01,