Transmission Media-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

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

2011/2012

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CHAPTER 7
Transmission Media
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The transmission media is located beneath the physical layer and controlled by
the physical layer.
3. Guided media have physical boundaries, while unguided media are unbounded.
5. Twisting ensures that both wires are equally, but inversely, affected by external
influences such as noise.
7. The inner core of an optical fiber is surrounded by cladding. The core is denser
than the cladding, so a light beam traveling through the core is reflected at the
boundary between the core and the cladding if the incident angle is more than the
critical angle.
9. In sky propagation radio waves radiate upward into the ionosphere and are then
reflected back to earth. In line-of-sight propagation signals are transmitted in a
straight line from antenna to antenna.
Exercises
11. See Table 7.1 (the values are approximate).
13. We can use Table 7.1 to find the power for different frequencies:
Tab le 7.1 Solution to Exercise 11
Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz
1 Km โˆ’3โˆ’5โˆ’7
10 Km โˆ’30 โˆ’50 โˆ’70
15 Km โˆ’45 โˆ’75 โˆ’105
20 Km โˆ’60 โˆ’100 โˆ’140
1 KHz dB = โˆ’3P
2 = P1 ร—10โˆ’3/10 = 100.23 mw
10 KHz dB = โˆ’5P
2 = P1 ร—10โˆ’5/10 = 63.25 mw
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CHAPTER 7

Transmission Media

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. The transmission media is located beneath the physical layer and controlled by the physical layer.
  2. Guided media have physical boundaries, while unguided media are unbounded.
  3. Twisting ensures that both wires are equally, but inversely , affected by external influences such as noise.
  4. The inner core of an optical fiber is surrounded by cladding. The core is denser than the cladding, so a light beam traveling through the core is reflected at the boundary between the core and the cladding if the incident angle is more than the critical angle.
  5. In sky propagation radio waves radiate upward into the ionosphere and are then reflected back to earth. In line-of-sight propagation signals are transmitted in a straight line from antenna to antenna.

Exercises

  1. See Table 7.1 (the values are approximate).
  2. We can use Table 7.1 to find the power for different frequencies:

Table 7.1 Solution to Exercise 11

Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km โˆ’ 3 โˆ’ 5 โˆ’ 7 10 Km โˆ’ 30 โˆ’ 50 โˆ’ 70 15 Km โˆ’ 45 โˆ’ 75 โˆ’ 105 20 Km โˆ’ 60 โˆ’ 100 โˆ’ 140

1 KHz dB = โˆ’ 3 P 2 = P 1 ร— 10 โˆ’3/10^ = 100.23 mw 10 KHz dB = โˆ’ 5 P 2 = P 1 ร— 10 โˆ’5/10^ = 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.

  1. We first make Table 7.2 from Figure 7.9 (in the textbook).

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 โˆ’50 dB (loss), then we can give the following listing of distance versus bandwidth.

  1. We can use the formula f = c / ฮป to find the corresponding frequency for each wave length as shown below (c is the speed of propagation): a. B = [(2 ร— 10 8 )/1000ร— 10 โˆ’9] โˆ’ [(2 ร— 10 8 )/ 1200 ร— 10 โˆ’9] = 33 THz b. B = [(2 ร— 10 8 )/1000ร— 10 โˆ’9] โˆ’ [(2 ร— 10 8 )/ 1400 ร— 10 โˆ’9] = 57 THz
  2. See Table 7.3 (The values are approximate).
  3. See Figure 7.1.

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 = โˆ’ 7 P 2 = P 1 ร— 10 โˆ’7/10^ = 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 โˆ’ 3 โˆ’ 7 โˆ’ 20 10 Km โˆ’ 30 โˆ’ 70 โˆ’ 200 15 Km โˆ’ 45 โˆ’ 105 โˆ’ 300 20 Km โˆ’ 60 โˆ’ 140 โˆ’ 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 โˆ’ 3 โˆ’ 1.1 โˆ’ 0. 10 Km โˆ’ 30 โˆ’ 11 โˆ’ 5 15 Km โˆ’ 45 โˆ’ 16.5 โˆ’ 7. 20 Km โˆ’ 60 โˆ’ 22 โˆ’ 10