Digital Transmission-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: Digital, Transmission, Scrambling, Unipolar, polar, Bipolar, Multilevel, Multitranstion, Scrabling, Parallel

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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CHAPTER 4
Digital Transmission
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The three different techniques described in this chapter are line coding, block cod-
ing, and scrambling.
3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits
per second (bps). The signal rate is the number of signal elements sent in 1s. The
unit is the baud.
5. When the voltage level in a digital signal is constant for a while, the spectrum cre-
ates very low frequencies, called DC components, that present problems for a sys-
tem that cannot pass low frequencies.
7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitran-
sition coding.
9. Scrambling, as discussed in this chapter, is a technique that substitutes long zero-
level pulses with a combination of other levels without increasing the number of
bits.
11. In parallel transmission we send data several bits at a time. In serial transmission
we send data one bit at a time.
Exercises
13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2.
a. r = 1 s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud
b. r = 1/2 s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud
c. r = 2 s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaud
d. r = 4/3 s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud
15. See Figure 4.1 Bandwidth is proportional to (3/8)N which is within the range in
Table 4.1 (B = 0 to N) for the NRZ-L scheme.
17. See Figure 4.2. Bandwidth is proportional to (12.5 / 8) N which is within the range
in Table 4.1 (B = N to B = 2N) for the Manchester scheme.
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CHAPTER 4

Digital Transmission

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. The three different techniques described in this chapter are line coding , block cod- ing , and scrambling.
  2. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps). The signal rate is the number of signal elements sent in 1s. The unit is the baud.
  3. When the voltage level in a digital signal is constant for a while, the spectrum cre- ates very low frequencies, called DC components , that present problems for a sys- tem that cannot pass low frequencies.
  4. In this chapter, we introduced unipolar , polar , bipolar , multilevel , and multitran- sition coding.
  5. Scrambling, as discussed in this chapter, is a technique that substitutes long zero- level pulses with a combination of other levels without increasing the number of bits.
  6. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time.

Exercises

  1. We use the formula s = c × N × (1/r) for each case. We let c = 1/2. a. r = 1 → s = (1/2) × (1 Mbps) × 1/ 1 = 500 kbaud b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/( 1/2 ) = 1 Mbaud c. r = 2 → s = (1/2) × (1 Mbps) × 1/ 2 = 250 Kbaud d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/( 4/3 ) = 375 Kbaud
  2. See Figure 4.1 Bandwidth is proportional to (3/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-L scheme.
  3. See Figure 4.2. Bandwidth is proportional to (12.5 / 8) N which is within the range in Table 4.1 (B = N to B = 2N) for the Manchester scheme.
  1. See Figure 4.3. B is proportional to (5.25 / 16) N which is inside range in Table 4. (B = 0 to N/2) for 2B/1Q.
  2. The data stream can be found as a. NRZ-I: 10011001. b. Differential Manchester: 11000100. c. AMI: 01110001.
  3. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N. We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 = 0 → P = 0. b. f /N = 50/100 = 1/2 → P = 0. c. f /N = 100/100 = 1 → P = 0. d. f /N = 150/100 = 1.5 → P = 0.

Figure 4.1 Solution to Exercise 15

Figure 4.2 Solution to Exercise 17

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

Case a

Case b

Case c

Case d

Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8 B (3 / 8) N

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

Case a

Case b

Case c

Case d

Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8 B (12.5 / 8) N