Data Communication and Networking - 4th edition exercise solution by Behrouz A Forouzan,

CHAPTER 7
Transmission Media

Solutions to 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:
Table 7.1 Solution to Exercise 11
Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz
1 Km −3 −5 −7
10 Km −30 −50 −70
15 Km −45 −75 −105
20 Km −60 −100 −140
1 KHz dB = −3 P2 = P1 ×10−3/10 = 100.23 mw
10 KHz dB = −5 P2 = P1 ×10−5/10 = 63.25 mw
2
The table shows that the power for 100 KHz is reduced almost 5 times, which may
not be acceptable for some applications.
15. 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
−50 dB (loss), then we can give the following listing of distance versus bandwidth.
17. We can use the formula f = c / λ to find the corresponding frequency for each wave
length as shown below (c is the speed of propagation):
a. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1200 × 10−9] = 33 THz
b. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1400 × 10−9] = 57 THz
19. See Table 7.3 (The values are approximate).
21. 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 = −7 P2 = P1 ×10−7/10 = 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 −3 −7 −20
10 Km −30 −70 −200
15 Km −45 −105 −300
20 Km −60 −140 −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 −3 −1.1 −0.5
10 Km −30 −11 −5
15 Km −45 −16.5 −7.5
20 Km −60 −22 −10
3
c. The incident angle (80 degrees) is greater than the critical angle (60 degrees).
We have reflection. The light ray returns back to the more dense medium.
Figure 7.1 Solution to Exercise 21
Critical angle = 60
Critical angle = 60
Critical angle = 60
Refraction
b. 60 degrees
Reflection
c. 80 degrees
Critical angle
Critical angle
a. 40 degrees
Refraction
Critical angle