360
8 Digital Optical Fiber Links
(b) Show that for an ideal OOK homodyne system, one needs 36 photons per
pulse to achieve a 10
−9 BER.
Answers to Selected Problems
8.1 The spaces in the following answer are inserted for clarity purposes only:
(a) Original code: 010 001 111 111 101 000 000 001 111 110
3B4B encoded: 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100
(b) The maximum number of consecutive identical bits is three.
8.2 (a) Original code: 0101 1101 0010 1110 1010 0111
4B5B encoded: 01011 11011 10100 11100 10110 01111
8.3 For system1, L = 12 km; for system 2, L = 11.3 km
8.4 (a) For the pin receiver L = 4.25 km; (b) For the APD receiver L = 7.0 km
8.7 (a) t sys = 4.90 ns; t sys < 0.7 Tb = 7.78 ns (b) t sys = 5.85 ns
8.9 12.1 dB at 622 Mb/s; 4.1 dB at 2.5 Gb/s
8.10
Information word
Code word
b1
b2
b3
b4
b1
b2
b3
b4
b5
b6
b7
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
1
1
1
0
0
1
0
0
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
1
0
0
1
0
0
0
1
0
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
0
1
1
0
0
1
1
0
1
1
0
0
1
1
1
0
1
1
1
0
0
1
1
0
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
0
1
0
0
1
1
0
1
0
1
0
1
0
0
1
1
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
1
1
0
1
1
1
0
1
0
1
0
1
1
1
0
1
1
1
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
8.11 (a) 110001011; (b) x
16
+ x
13
+ x
12
+ x
10
+ x
9
+ x
8
+ x
7
+ x
5
+ x
4
+ x
2
+
1
8.16 i IF (t) = 0.67 μA
8.17 (a) λ = 5.6 × 10
–6 nm; (b) λ = 1.0 × 10
–4 nm.
8 Digital Optical Fiber Links
(b) Show that for an ideal OOK homodyne system, one needs 36 photons per
pulse to achieve a 10
−9 BER.
Answers to Selected Problems
8.1 The spaces in the following answer are inserted for clarity purposes only:
(a) Original code: 010 001 111 111 101 000 000 001 111 110
3B4B encoded: 0101 0011 1011 0100 1010 0010 1101 0011 1011 1100
(b) The maximum number of consecutive identical bits is three.
8.2 (a) Original code: 0101 1101 0010 1110 1010 0111
4B5B encoded: 01011 11011 10100 11100 10110 01111
8.3 For system1, L = 12 km; for system 2, L = 11.3 km
8.4 (a) For the pin receiver L = 4.25 km; (b) For the APD receiver L = 7.0 km
8.7 (a) t sys = 4.90 ns; t sys < 0.7 Tb = 7.78 ns (b) t sys = 5.85 ns
8.9 12.1 dB at 622 Mb/s; 4.1 dB at 2.5 Gb/s
8.10
Information word
Code word
b1
b2
b3
b4
b1
b2
b3
b4
b5
b6
b7
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
1
1
1
0
0
1
0
0
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
1
0
0
1
0
0
0
1
0
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
0
1
1
0
0
1
1
0
1
1
0
0
1
1
1
0
1
1
1
0
0
1
1
0
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
0
1
0
0
1
1
0
1
0
1
0
1
0
0
1
1
1
0
1
1
1
0
1
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
1
1
0
1
1
1
0
1
0
1
0
1
1
1
0
1
1
1
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
8.11 (a) 110001011; (b) x
16
+ x
13
+ x
12
+ x
10
+ x
9
+ x
8
+ x
7
+ x
5
+ x
4
+ x
2
+
1
8.16 i IF (t) = 0.67 μA
8.17 (a) λ = 5.6 × 10
–6 nm; (b) λ = 1.0 × 10
–4 nm.
