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This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: WWW, HTTP, Proxy, Server, Web, Document, Active, Static, Head, Copy, Delete, Modified, Authentication
Typology: Exercises
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Table 7.1 Solution to Exercise 11
Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km โ 3 โ 5 โ 7 10 Km โ 30 โ 50 โ 70 15 Km โ 45 โ 75 โ 105 20 Km โ 60 โ 100 โ 140
1 KHz dB = โ 3 P 2 = P 1 ร 10 โ3/10^ = 100.23 mw 10 KHz dB = โ 5 P 2 = P 1 ร 10 โ5/10^ = 63.25 mw
The table shows that the power for 100 KHz is reduced almost 5 times, which may not be acceptable for some applications.
If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance. For example, if we can tolerate a maximum attenuation of โ50 dB (loss), then we can give the following listing of distance versus bandwidth.
a. The incident angle (40 degrees) is smaller than the critical angle (60 degrees). We have refraction .The light ray enters into the less dense medium. b. The incident angle (60 degrees) is the same as the critical angle (60 degrees). We have refraction. The light ray travels along the interface.
100 KHz dB = โ 7 P 2 = P 1 ร 10 โ7/10^ = 39.90 mw
Table 7.2 Solution to Exercise 15
Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km โ 3 โ 7 โ 20 10 Km โ 30 โ 70 โ 200 15 Km โ 45 โ 105 โ 300 20 Km โ 60 โ 140 โ 400
Distance Bandwidth 1 Km 100 KHz 10 Km 1 KHz 15 Km 1 KHz 20 Km 0 KHz
Table 7.3 Solution to Exercise 19
Distance dB at 800 nm dB at 1000 nm dB at 1200 nm 1 Km โ 3 โ 1.1 โ 0. 10 Km โ 30 โ 11 โ 5 15 Km โ 45 โ 16.5 โ 7. 20 Km โ 60 โ 22 โ 10